NCERT Solutions for Maths Class 10 Chapter 8

Exercise 8.1

Question 1: In \(ΔABC\) right angled at \(B, AB = 24 cm, BC = 7 m\). Determine

(i) sin A, cos A

(ii) sin C, cos C

Answer:  Applying Pythagoras theorem for \(ΔABC\), we obtain \(AC^2 = AB^2  + BC^2\)

\(=(24cm)^2+(7cm)^2\)

\(=(567+49)cm^2\)

\(=625cm^2\)

\(∴AC=\sqrt{625}cm=25cm\)

(i) 

\(Sin A=\frac{Side\space opposite\space to\angle A}{Hypotenuse}=\frac{BC}{AC}\)

\(=\frac{7}{25}\)

\(cos A=\frac{Side\space adjacent\space to \angle A}{Hypotenuse}=\frac{AB}{AC}=\frac{24}{25}\)

(ii) 

\(Sin\space C=\frac{Side\space opposite\space to\angle C}{Hypoenuse}=\frac{AB}{AC}\)

\(=\frac{24}{25}\)

\(cos\space C=\frac{Side\space adjacent\space to \angle C}{Hypotenuse}=\frac{BC}{AC}\)

\(=\frac{7}{25}\)

Question 2: In the given figure dind \(tan\space P-cot\space R\)

Answer: 

Applying Pythagoras theorem for \(ΔPQR\), we obtain \(PR^2  = PQ^2  + QR^2\)  

\((13 cm)^2  = (12 cm)^2  + QR^2\)  

\(169 cm^2=144 cm^2+QR^2\)

\(25 cm^2=QR^2\)

\(QR=5cm\)

\(tan\space P=\frac{Side\space opposite\space to\space\angle P}{Side\space adjuacent\space to\space \angle P}\)

\(=\frac{QR}{PQ}\)\(=\frac{5}{12}\)

\(cot\space P=\frac{Side\space adjacent\space to\space\angle R}{Side\space opposite\space to\space\angle R}\)

\(=\frac{QR}{PQ}\)\(=\frac{5}{12}\)

\(tan\space P-cot\space R=\frac{5}{12}-\frac{5}{12}=0\)

Question 3: If \(sin\space A\frac{3}{4}\), calculate \(cos\space A\) and \(tan\space A\)

Answer: 

Let \(ΔABC\) be a right-angled triangle, right-angled at point \(B\).

Given that,

\(sin\space A=\frac{3}{4}\)

\(\frac{BC}{AC}=\frac{3}{4}\)

Let \(BC\) be \(3k\).

Therefore, \(AC\) will be \(4k\), where k is a positive integer.

Applying Pythagoras theorem in \(ΔABC\), we obtain

\(AC^2 = AB^2 + BC^2

\((4k)^2 = AB^2 + (3k)^2\)
\(16k^2 − 9k^2 = AB^2\)
\(7k^2 = AB^2\)

\(AB =\sqrt{7}k\)

\(cos\space A=\frac{Side\space adjacent\space to\space\angle A}{Hypotenuse}\)

\(\frac{AB}{AC}=\frac{\sqrt7k}{4k}=\frac{\sqrt7}{4}\)

\(tan\space A=\frac{Side\space opposite\space to\space\angle A}{Side\space adjacent\space to\space\angle A}\)

\(=\frac{BC}{AB}=\frac{3k}{\sqrt7k}=\frac{3}{\sqrt7}\)

Question 4: Given \(15\space cot\space A = 8\). Find \(sin\space A\) and \(sec\space A\)

Answer: 

Consider a right-angled triangle, right-angled at \(B).

\(cot\space=\frac{Side\space adjucent\space to\space \angle A}{Side\space opposite\space to\space \angle A}\)

\(=\frac{AB}{BC}\)

It is given that,

\(cot\space A=\frac{8}{15}\)

\(\frac{AB}{BC}=\frac{8}{15}\)

Let \(AB\) be \(8k\).

Therefore, \(BC\) will be \(15K\), where is a postive integer.

Applying Pyhagoras theorem in \(\Delta ABC\), we obtain

\(AC^2=AB^2+BC^2\)

\(=(8k)^2+(15k)^2\)

\(=64k^2+225k^2\)

\(=289k^2\)

\(AC=17k\)

\(sin\space A=\frac{Hypotenuse}{Side\space adhacent\space to\space\angle A}\)

\(=\frac{AC}{AB}=\frac{17}{8}\)

Question 5: Given \(sec\space\theta=\frac{13}{12}\), calculate all other triogonometric ratios.

Answer:

Consider a right-angle triangle \(\Delta ABC\), right-angled at point \(B\).

\(sec\space\theta=\frac{Hypotenuse}{Side\space adjacent\space to\space\angle\theta}\)

\(\frac{13}{12}=\frac{AC}{AB}\)

If \(AC\) is \(13k\), AB will be \(12k\), where \(k\) is a positive integer.

Applying Pythagoras theorem in \(\Delta ABC\), we obtain

\((AC)^2=(AB)^2+(BC)^2\)

\((13k)^2=(12k)^2+(BC)^2\)

\(169k^2=144k^2+BC^2\)

\(25k^2=BC^2\)

\(BC=5k\)

\(sin\space\theta=\frac{Side\space opposite\space to\space\angle\theta}{Hypotenuse}=\frac{BC}{AC}=\frac{5k}{13k}=\frac{5}{13}\)

\(cos\space\theta=\frac{Side\space adjacent\space to\space\angle\theta}{Hypotenuse}=\frac{AB}{AC}=\frac{12k}{13k}=\frac{12}{13}\)

\(tan\space\theta=\frac{Side\space opposite\space to\space\angle\theta}{Side\space adjacent\space to\angle\theta}=\frac{BC}{AC}=\frac{5k}{12k}=\frac{5}{12}\)

\(cot\space\theta=\frac{Side\space adjacent\space to\angle\theta}{Side\space opposite\space to \angle\theta}=\frac{AB}{BC}=\frac{12k}{5k}=\frac{12}{5}\)

\(cosec\space\theta=frac{Hypotenuse}{Side\space opposite\space to\angle\theta}=\frac{AC}{BC}=\frac{13k}{5k}=\frac{13}{5}\)

Question 6: If \(∠A\) and \(∠B\) are acute angles such that \(cos\space A = cos\space B\), then show that \(∠A = ∠B\).

Answer:

Let us consider a triangle \(ABC\) in which \(CD ⊥ AB\).

It is given that

\(cos\space A=cos\space B\)

\(⇒\frac{AD}{AC}=\frac{BD}{BC}\)      …(i)

We have to prove \(\angle A=\angle B\) to prove this,

Let us extend \(AC\) to \(P\) such that \(BC=CP\).

From equation (i), we obtain

\(\frac{AD}{BC}=\frac{AC}{BC}\)

\(⇒\frac{AD}{BD}=\frac{AC}{CP}\)  (By construction, we have \(BC=CP\))   ..(ii)

By using the converse of \(B.P.T\),

\(CD\parallel BP\)

\(⇒\angle ACD=\angle CPB\)  (Corresponding angles)   …(iii)

And \(\angle BCD=\angle CBP\)  (Altenate interior angles) ….(iv)

By construction, we have \(BC = CP\).

\(∴  ∠CBP = ∠CPB\) (Angle opposite to equal sides of a triangle) … (v)

From equations (iii), (iv), and (v), we obtain

\(∠ACD = ∠BCD\) … (vi)

In \(ΔCAD\) and \(ΔCBD\),

\(∠ACD = ∠BCD\) [Using equation (vi)]
\(∠CDA = ∠CDB\) [Both \(90^0\)]

Therefore, the remaining angles should be equal.

\(∴∠CAD = ∠CBD\)

\(⇒  ∠A = ∠B\)

Alternatively,

Let us consider a triangle \(ABC\) in which \(CD ⊥ AB\).

It is given that,

\(cos\space A = cos\space B\)

\(⇒\frac{AD}{AC}=\frac{BD}{BC}\)

\(⇒\frac{AD}{BD}=\frac{AC}{BC}\)

Let \(\frac{AD}{BD}=\frac{AC}{BC}=k\)

\(⇒AD=k\space BD\)   …(i)

And \(AC=k\space BC\) … (ii)

\(⇒ (k\space BC)^2 − (k\space BD)^2 = BC^2 − BD^2\)
\(⇒ k^2 (BC^2 − BD^2) = BC^2 − BD^2\)
\(⇒ k^2 = 1\)

\(⇒ k = 1\)

Putting this value in equation (ii), we obtain \(AC = BC\)

\(⇒ ∠A = ∠B\)   (Angles opposite to equal sides of a triangle)

Question 7: If \(cot\space\theta=\frac{7}{8}\), evalute

(i) \(\frac{(1+sin\theta)(1-sin\theta)}{(1+cos\theta)(1-cos\theta)}\)

(ii) \(cot\space2\space \theta\)

Answer:

Let us consider a right triangle \(ABC\), right-angled at point \(B\).

\(cot\space\theta=\frac{Side\space adjacent\space\to\space\angle\theta}{Side\space opposite\space to\angle\theta}=\frac{BC}{AB}\)

\(=\frac{7}{8}\)

If

\(BC\) is \(7k\), then \(AB\) will be \(8k\), where \(k\) is a positive integer.

Applying Pythagoras theorem in \(ΔABC\), we obtain
\(AC^2 = AB^2 + BC^2\)

\(= (8k)^2 + (7k)^2\)
\(= 64k^2 + 49k^2\)
\(= 113k^2\)

\(AC =\sqrt{113}k\)

\(sin\space\theta=\frac{Side\space opposite\space to\space\angle\theta}{Hypotenuse}=\frac{BA}{AC}\)

\(=\frac{8k}{\sqrt{113k}}=\frac{8}{\sqrt{113}}\)

\(cos\space\theta=\frac{Side\space adjacent\space to\space\angle\theta}{Hypoyenuse}=\frac{BA}{AC}\)

\(=\frac{7}{\sqrt{113k}}=\frac{7}{\sqrt{113}}\)

\(\frac{(1+sin\space\theta)(1-sin\space\theta)}{(a+cos\space\theta)(1-cos\space\theta)}=\frac{(1-sin^2\space\theta)}{(1-cos^2\space\theta)}\)

(i)

\(\frac{1(\frac{8}{\sqrt{113}})^2}{1-(\frac{7}{\sqrt{113}})^2}=\frac{1-\frac{64}{113}}{1-\frac{49}{113}}\)

\(=\frac{1(\frac{8}{\sqrt{113}})^2}{1-(\frac{7}{\sqrt{113}})^2}=\frac{49}{64}\)

(ii) \(cot^2\space\theta=(cot\space\theta)^2=(\frac{7}{8})^2=\frac{49}{64}\)

Question 8: If \(3\space cot\space A=4\), Check whether \(\frac{1-tan^2\space A}{1+tan^2\space A}=cos^2\space A-sin^2\) or not.

Answer:

It is given that \(3\space cot\space A=4\)

Or \(cot\space A=\frac{4}{3}\)

Consider a right triangle \(ABC\), right-angled at point \(B\).

If \(AB\) is \(4k\), then \(BC\) will be \(3k\), where k is a positive integer. In \(ΔABC\),

\((AC)^2 = (AB)^2 + (BC)^2\)
\(= (4k)^2 + (3k)^2\)

\(= 16k^2 + 9k^2\)
\(= 25k^2\)

\(AC = 5k\)

\(cos\space A=\frac{Side\space adjacent\space to\space\angle A}{Hypotenuse}=\frac{AB}{AC}\)

\(=\frac{4k}{5}=\frac{4}{5}\)

\(sin\space A=\frac{Side\space opposite\space to\space\angle A}{Hypotenuse}=\frac{BC}{AB}\)

\(=\frac{3k}{5k}=\frac{3}{5}\)

\(tan\space A=\frac{Side\space opposite\space to\space\angle A}{Hypotenuse}=\frac{BC}{AB}\)

\(=\frac{3k}{4k}=\frac{3}{4}\)

\(\frac{1-tan^2\space A}{1+tan^2\space A}=\frac{1-(\frac{3}{4})^2}{1+(\frac{3}{4})^2}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}\)

\(=\frac{\frac{7}{16}}{\frac{25}{16}}=\frac{7}{25}\)

\(cos^2\space A-sin^2\space A=(\frac{4}{5})^2-(\frac{3}{5})^2\)

\(=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}\)

\(∴\frac{1-tan^2\space A}{1+tan^2\space A}=cos^2\space A-sin^2\space A\)

Question 9: In \(ΔABC\), right angled at \(B\).  If \(tan\space A=\frac{1}{\sqrt3}\), find the value of

(i) sin A cos C + cos A sin C
(ii) cos A cos C − sin A sin C

Answer:

\(tan\space A=\frac{1}{\sqrt3}\)

\(\frac{BC}{AB}=\frac{1}{\sqrt3}\)

If \(BC\) is \(k\), then \(AB\) will be \(\sqrt3k\), where \(k\) is a positive integer.

In \(\Delta ABC\),

\(AC^2=AB^2+BC^2\)

\(=(\sqrt3k)^2+9k)^2\)

\(=3k^2=4k^2\)

\(∴AC=2k\)

\(sin\space A=\frac{Side\space opposite\space to\space\angle A}{Hypotenuse}=\frac{BC}{AC}=\frac{k}{2k}=\frac{1}{2}\)

\(cos\space A=\frac{Side\space adjacent\space to\space\angle C}{Hypotenuse}=\frac{AB}{AC}=\frac{\sqrt3k}{2k}=\frac{\sqrt3}{2}\)

\(sin\space C=\frac{Side\space opposite\space to\space\angle C}{Hypotenuse}=\frac{\sqrt3k}{2k}=\frac{1}{2}\)

\(cos\space C=\frac{Side\space adjacent\space to\space\angle C}{Hypotenuse}=\frac{BC}{AC}=\frac{k}{2k}=\frac{1}{2}\)

(i) sin A cos C + cos A sin C

\(=(\frac{1}{2})(\frac{1}{2})+(\frac{\sqrt3}{2})(\frac{\sqrt3}{2})=\frac{1}{4}+\frac{3}{4}\)

\(=\frac{4}{4}=1\)

(ii) cos A cos C – sin A sin C

\(=(\frac{\sqrt3}{2})(\frac{1}{2})-(\frac{1}{2})(\frac{\sqrt3}{2})=(\frac{\sqrt3}{4}-\frac{\sqrt3}{4}=0\)

Question 10: In \(ΔPQR\), right angled at \(Q,\space PR + QR = 25\space cm\) and \(PQ = 5\space cm\). Determine the values of \(sin\space P,\space cos\space P\) and \(tan\space P\).

Answer:

Given that, \(PR + QR = 25\)
\(PQ = 5\)

Let \(PR\) be \(9x\).

Therefore, \(QR = 25 − x\)

Applying Pythagoras theorem in \(ΔPQR\), we obtain

\(PR^2 = PQ^2 + QR^2\)

\(x^2 = (5)^2 + (25 − x)^2\)

\(x^ = 25 + 625 + x^2 − 50x\)
\(50x = 650\)

\(x = 13\)

Therefore, \(PR= 13\space cm\)

\(QR = (25 − 13)\space cm = 12\space cm\)

\(sin\space P=\frac{Side\space opposite\space to\space\angle P}{Hypotenuse}=\frac{QR}{PR}=\frac{12}{13}\)

\(cos\space P=\frac{Side\space adjacent\space to\space\angle P}{Hypotenuse}=\frac{PQ}{PR}=\frac{5}{13}\)

\(tan\space P=\frac{Side\space opposite\space to\space\angle P}{Side\space adjacent\space to\space\angle P}=\frac{QR}{PQ}=\frac{12}{5}\)

Question 11: State whether the following are true or false. Justify your answer.

(i) The value of \(tan\space A\) is always less than \(1\).

(ii) \(sec\space A =\frac{4}{3}\)  for some value of angle \(A\).

(iii) \(cos\space A\) is the abbreviation used for the cosecant of angle \(A\).

(iv) \(cot\space A\) is the product of cot and \(A\)

(v) \(sin\space θ =\frac{4}{3}\), for some angle \(θ\)

Answer:

(i) Consider a \(\Delta ABC\), right-angled at \(B\)

\(tan\space A\frac{Side\space opposite\space to\space\angle A}{Side\space adjacent\space to\space\angle A}\)

\(\frac{12}{5}\)

But \(\frac{12}{5}\gt1\)

\(∴tan\space A\gt1\)

So, \(tan\space A\gt1\) is not always true.

Hence, the given statement is false.

(ii) \(sec\space A=frac{12}{5}\)

\(\frac{Hypotenuse}{Side\space adjacent\space to\space\angle A}=\frac{12}{5}\)

\(\frac{AC}{AB}=\frac{12}{5}\)

Let \(AC\) be \(12k,\space AB\) will be \(5k\), where \(k\) is a positive integer.

Applying Pythagoras theorem in \(ΔABC\), we obtain
\(AC^2 = AB^2 + BC^2\)

\((12k)^2 = (5k)^2 + BC^2\)
\(144k^2 = 25k^2 + BC^2\)
\(BC^2 = 119k^2\)

\(BC = 10.9k\)

It can be observed that for given two sides \(AC = 12k\) and \(AB = 5k\),

\(BC\) should be such that,

\(AC − AB < BC < AC + AB\)
\(12k − 5k < BC < 12k + 5k\)
\(7k < BC < 17 k\)

However, \(BC = 10.9k\). Clearly, such a triangle is possible and hence, such value of sec \(A\) is possible.

Hence, the given statement is true.

(iii) Abbreviation used for cosecant of angle \(A\) is \(cosec\space A\). And \(cos\space A\) is the abbreviation used for cosine of \(∠A\).

Hence, the given statement is false.

(iv) \(cot\space A\) is not the product of cot and \(A\). It is the cotangent of \(∠A\). Hence, the given statement is false.

(v) \(sin\space\theta=\frac{4}{3}\)

We know that in a right-angled triangle,

\(sin\space\theta=\frac{Side\space opposite\space to\space\angle\theta}{Hypotenuse}\)

In a right-angled triangle, hypotenuse is always greater than the remaining two sides.

Therefore, such value of sin \(θ\) is not possible. Hence, the given statement is false

Exercise 8.2

Question 1: Evaluate the following

(i) \(sin\space60^0\space cos30^0+sin\space 30^0\space cose\space 60^0\)

(ii) \(2\space tan^2 \space45^0+cos^2\space 30-sin^2\space60^0\)

(i) \(sin\space60^0\space cos30^0+sin\space 30^0\space cose\space 60^0\)

(ii) \(2\space tan^2 \space45^0+cos^2\space 30-sin^2\space60^0\)

(iii) \(\frac{cos\space45^0}{sec\space30^0+cos\space60^0+cot\space45^0}\)

(iv) \(\frac{sin\space30^0+tan\space45^0-cosec\space60^0}{sec\space30^0+cos\space60^0+cot\space45^0}\)

(v) \(\frac{5\space cos^2\space 60^0+4\space sec^2\space 30^0-tan^2\space 45^0}{sin^2\space 30^0+cos\space 30^0}\)

Answer:

(i) \(sin\space60^0\space cose\space 30^0+sin\space30^0\space cos60^0\)

\(=(\frac{\sqrt3}{2})(\frac{\sqrt3}{2}+(\frac{1}{2}(\frac{1}{2})\)

\(=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1\)

(ii) \(2\space tan^2\space 45^0+cos62\space30^0-sin\space^2 60^0\)

\(=2(1)^2+(\frac{\sqrt3}{2})^2-(\frac{\sqrt3}{2})^2\)

\(=2+\frac{3}{4}-\frac{3}{4}=2\)

(iii) \(\frac{cos\space45^0}{sec\space30^0+cosec\space30^0}\)

\(=\frac{\frac{1}{\sqrt2}}{\frac{2}{\sqrt3}}+2=\frac{\frac{1}{\sqrt2}}{\frac{2+2\sqrt3}{\sqrt3}}\)

\(=\frac{\sqrt3}{\sqrt2(2_2\sqrt3)}=\frac{\sqrt3}{\sqrt(2+2\sqrt3)}=\frac{\sqrt3}{2\sqrt2+2\sqrt6}\)

\(=\frac{\sqrt3(2\sqrt6-2\sqrt2}{(2\sqrt6=2\sqrt2)(2\sqrt6-2\sqrt6}\)
\(=\frac{2\sqrt3(\sqrt6-\sqrt2)}{(2\sqrt6)^2-(2\sqrt2)^2}\)
\(=\frac{2\sqrt3(\sqrt6-\sqrt2)}{24-8}\)
\(=\frac{2\sqrt3(\sqrt6-\sqrt2)}{16}\)
\(=\frac{\sqrt18-\sqrt6}{8}\)
\(=\frac{3\sqrt2-\sqrt6}{8}\)

(iv) \(\frac{sin\space30^0+tan\space45^0-cosec\space60^0}{sec\space30^+cos\space60^0=cot\space45^0}\)

\(=\frac{\frac{1}{2}+1-\frac{2}{\sqrt3}}{\frac{2}{\sqrt3}+\frac{1}{2}+1}=\frac{\frac{3}{2}-\frac{2}{\sqrt3}}{\frac{3}{2}+\frac{2}{\sqrt3}}\)

\(=\frac{\frac{3\sqrt3-4}{2\sqrt3}}{\frac{3\sqrt3+4}{2\sqrt3}}+\frac{(3\sqrt3-4)}{(3\sqrt3+4)}\)

\(=\frac{(3\sqrt3-4)(3\sqrt3-4)}{(3\sqrt3+4)(3\sqrt3-4}=\frac{(3\sqrt3-4)^2}{(3\sqrt3)^2-(4)^2}\)

\(=\frac{27+16-24\sqrt3}{27-16}=\frac{43-24\sqrt3}{11}\)

(v) \(\frac{5\space cose^2\space60^0=4\space sec^2\space30^0-tan^2\space45^0}{sin^2\space30^0+cos^2\space30^0}\)

\(=\frac{5(\frac{1}{2})^2)+4(\frac{2}{\sqrt3})^2-(1)^2}{(\frac{1}{2})^2+(\frac{\sqrt3}{2})^2}\)

\(=\frac{5(\frac{1}{4})+(\frac{16}{3})-1}{\frac{1}{4}+\frac{3}{4}}\)

\(=\frac{\frac{15+64+-12}{12}}{\frac{4}{4}}=\frac{67}{12}\)

Question 2: Choose the correct option and justify your choice.

(i) \(\frac{2\space tan\space 30^0}{1=tan^2\space 30^0}=?\)

(a) \(sin\space60^0\)

(b) \(cos\space 60^0\)

(c) \(tan\space 60^0\)

(d) \(sin\space 30^0\)

Answer: 

(i) \(\frac{2\space tan\space 30^0}{1=tan^2\space30^0}\)

\(=\frac{2(\frac{1}{\sqrt3})}{1+(\frac{1}{\sqrt3})^2}=\frac{\frac{2}{\sqrt3}}{1+\frac{1}{3}}=\frac{\frac{2}{\sqrt3}}{\frac{4}{3}}\)

\(=\frac{6}{\sqrt3}=\frac{\sqrt3}{2}\)

Out of the given alternatives, only \(sin\space 60^0=\frac{\sqrt3}{2}\)

(ii) \(\frac{1-tan^2\space45^0}{1+tan^2\space 45^0}\)

(a) \(tan\space 90^0\)

(b) \(1\)

(c) \(sin\space 45^0\)

(d) \(0\)

Answer: 

(ii) \(\frac{1-tan^2\space45^0}{1+tan^2\space45^0}=?\)

\(=\frac{1-(1)^2}{1+(1)^2}=\frac{1-1}{1+1}=\frac{0}{2}=0\)

(iii) \(sin\space2A=2\space sin\space A\) is true when \(A=?\)

(a) \(0^0\)

(b) \(30^0\)

(c) \(45^0\)

(d) \(60^0\)

Answer: Out of the given alternatives, only \(A = 0^0\) is correct.

As \(sin\space 2A = sin\space 0^0 = 0\)

\(2\space sin\space A = 2\space sin\space 0^0 = 2(0) = 0\)

(iv) \(\frac{2\space tan\space30^0}{1-tan^2\space30^0}\)

(a) \(cos\space 60^0\)

(b) \(sin\space60^0\)

(c) \(tan\space60^0\)

(d) \(sin\space 30^0\)

Answer: \(\frac{2\space tan\space30^0}{1-tan^2\space 30^0}=?\)
\(=\frac{2(\frac{1}{\sqrt3})}{1-(\frac{1}{\sqrt3})^2}=\frac{\frac{2}{\sqrt3}}{1-\frac{1}{3}}=\frac{\frac{2}{\sqrt3}}{\frac{2}{3}}\)
\(=\sqrt3\)

Out of the given alternatives, only \(tan\space60^0=\sqrt3\)

Question 3: If tna \((A+B)=\sqrt3\) and \(tan\space9A-B)=\frac{1}{\sqrt3}\); \(0^0\lt A+B ≤90^0,A\lt B\) and \(A\) and \(B\).

Answer: \(tan \space (A+B)=\sqrt3\)

\(⇒tan\space(A+B)=tan\space⁡60\)

\(⇒A+B=60\)      ..(1)

\(tan\space(A-B)=\frac{1}{\sqrt3}\)

\(⇒ tan\space (A − B) = tan\space30\)
\(⇒ A − B = 30\) … (2)

On adding both equations, we obtain
\(2A = 90\)

\(⇒ A = 45\)

From equation (1), we obtain
\(45 + B = 60\)

\(B = 15\)

Therefore, \(∠A = 45^0\) and \(∠B = 15^0\)

Question 4: State whether the following are true or false. Justify your answer.

(i) \(sin \space (A + B) = sin\space A + sin\space B\)

(ii) The value of sin \(θ\) increases as \(θ \) increases

(iii) The value of cos \(θ\) increases as \(θ\) increases

(iv) \(sin\spaceθ = cos\spaceθ\) for all values of \(θ\)

(v) cot \(A\) is not defined for \(A = 0^0\)

(i) \(sin\space (A + B) = sin\space A + sin\space B\)
Let \(A = 30^0\) and \(B = 60^0\)
\(sin\space (A + B) = sin\space (30^0 + 60^0)\)
\(= sin\space 90^0\)

\(= 1\)

\(sin\space A + sin\space B = sin\space 30^0 + sin\space 60^0\)

\(=\frac{1}{2}+\frac{\sqrt3}{2}=\frac{1+\sqrt3}{2}\)

Clearly, \(sin\space (A + B) ≠ sin\space A + sin\space B\)
Hence, the given statement is false

(ii) The value of sin \(θ\) increases as \(θ\) increases in the interval of \(0^0 < θ < 90^0\) as sin \(0^0 = 0\)

\(sin\space30^0=\frac{1}{2}=0.5\)

\(sin\space45^0=\frac{1}{\sqrt2}=0.707\)

\(sin\space60^0=\frac{\sqrt3}{2}=0.866\)

\(sin\space90^0=1\)

Hence, the given statement is true.

(iii) \(cos\space0^0=1\)

\(cos\space30^0=\frac{\sqrt3}{2}=0.866\)

\(cos\space45^0=\frac{1}{\sqrt2}=0.707\)

\(cos\space60^0=\frac{1}{2}=0.5\)

\(cos\space90^0=0\)

It can be observed that the value of cos \(θ\) does not increase in the interval of \(0^0 < θ < 90^0\).

Hence, the given statement is false.

(iv) \(sin\space θ = cos\space θ\) for all values of \(θ\).
This is true when \(θ = 45^0\)

As \(sin\space45^0=\frac{1}{\sqrt2}\)

\(cos\space45^0=\frac{1}{\sqrt2}\)

It is not true for all other values of \(\theta\)

As \(sin\space30^0=\frac{1}{2}\) and \(cos\space30^0=\frac{\sqrt3}{2}\)

Hence, the given statement is false.

(v) \(cot\space A\) is not defined for \(A=0^0\)

As, \(cor\space A=\frac{cos\space A}{sin\space A}\)

\(cot\space 0^0=\frac{cos\space0^0}{sin\space0^0}=\frac{1}{0}=\) undefined

Hence, the given statement is true.

Exercise 8.3

Question 1:

Evaluate

(i) \(\frac{sin\space180^0}{cos\space72^0}\)

(ii) \(\frac{tan\space64^0}{cot\space64^0}\)

(iii) \(cos\space48^0-sin\space42^0\)

(iv) \(cosec\space31^0-sec\space59^0\)

Answer: 

(i) \(\frac{sin\space18^0}{cos\space72^0}=\frac{sin(90^0-72^0)}{cos\space72^0}\)

\(=\frac{cos\space72^0}{cos\space72^0}=1\)

(ii) \(\frac{tan\space26^0}{cot\space64^0}=\frac{tan(90^0-64^0)}{cot\space64^0}\)

\(=\frac{cot\space64^0}{cot\space64^0}=1\)

(iii) \(cos\space48^0-sin\space42^0=cos\space(90^0-42^0)-sin\space42^0\)

\(=sin\space42^0-sin\space42^0\)

\(=0\)

(iv) \(cos\space48^0-sin\space42^0=cos(90^0-42^0)-sin\space42^0\)

\(=sin\space42^0-sin\space42^0\)

\(=0\)

(iv) \(cosec\space31^0-sec\space59^0=cosec(90^0-59^0)-sec\space59^0\)

\(=sec\space59^0-sec\space59^0\)

\(=0\)

Question 2:

Show that

(i) \(tan\space48^0\space tan\space23^0\space tan\space42^0\space tan\space 67^0=1\)

(ii) \(cos\space38^0\space cos\space52^0-sin\space38^0\space sin\space 52^0=0\)

Answer:

(i) \(tan\space 48^0\space tan\space 23^0\space tan\space 42^0\space tan\space 67^0\)

\(= tan\space (90^0 − 42^0)\space tan (90^0 − 67^0)\space tan\space 42^0\space tan\space 67^0 = cot\space 42^0\space cot\space 67^0\space tan\space 42^0\space tan\space 67^0\)

\(= (cot\space 42^0\space tan\space 42^0)\space (cot\space 67^0\space tan\space 67^0)\)

\(= (1) (1)\)

\(= 1\)

(ii) \(cos\space 38^0\space cos\space 52^0 − sin\space 38^0\space sin\space 52^0\)

\(= cos\space (90^0 − 52^0)\space cos\space (90^0−38^0) − sin\space 38^0\space sin\space 52^0\)

\(= sin\space 52^0\space sin\space 38^0 − sin\space 38^0\space sin\space 52^0\)

\(= 0\)

Question 3: If \(tan\space 2A = cot\space (A− 18^0)\), where \(2A\) is an acute angle, find the value of \(A\).

Answer:

Given that,

\(tan\space 2A = cot\space (A− 18^0)\)

\(cot\space (90^0 − 2A) = cot\space (A −18^0)\)
\(90^0 − 2A = A− 18^0\)

\(108^0 = 3A\)

\(A = 36^0\)

Question 4: If \(tan\space A = cot\space B\), prove that \(A + B = 90^0\)

Answer:

Given that,

\(tan\space A = cot\space B\)

\(tan\space A = tan\space (90^0 − B)\)
\(A = 90^0 − B\)

\(A + B = 90^0\)

Question 5: If \(sec\space 4A = cosec\space (A− 20^0)\), where \(4A\) is an acute angle, find the value of \(A\).

Answer:

Given that,

\(sec\space 4A = cosec\space (A − 20^0)\)

\(cosec\space (90^0 − 4A) = cosec\space (A − 20^0)\)
\(90^0 − 4A= A− 20^0\)

\(110^0 = 5A\)

\(A = 22^0\)

Question 6: If \(A,B\) and \(C\) arre interior angles of a triangle of a triangle \(ABC\) then show that \(sin\space(\frac{B+C}{2})=cos\frac{A}{2}\)

Answer: We know that for a triangle \(ABC\),
\(∠ A + ∠B + ∠C = 180^0\)
\(∠B + ∠C= 180^0 − ∠A\)

\(\frac{∠B+∠C}{2}=90^0=-\frac{∠A}{2}\)

\(sin(\frac{B+C}{2})=sin(90^0-\frac{A}{2})\)

\(=cos(\frac{A}{2})\)

Question 7: Express sin \(67^0 + cos\space 75^0\) in terms of trigonometric ratios of angles between \(0^0\) and \(45^0\).

Answer:

\(sin\space67^0+cos\space75^0\)

\(=sin(90^0-23^0)+cos(90^0-15^0)\)

\(=cos\space23^0+sin\space15^0\)

Exercise 8.4

Question 1: Express the trigonometric ratios sin \(A,\space sec\space A\) and tan \(A\) in terms of \(cot\space A\).

Answer:

We know that,

\(cosec^2\space A=1+cot^2\space A\)

\(\frac{1}{cosec^2\space A}=\frac{1}{1+cot^2\space A}\)

\(sin^2\space A=\frac{1}{1+cot^2\space A}\)

\(\sqrt{11+cot^2\space A}\) will always be positive as we are two positive wuantireies.

Therfore, \(sin\space A=\frac{1}{\sqrt{1+cot^2\space A}}\)

We know that, \(tan\space A=\frac{sin\space A}{cos\space A}\)

Howevere, \(cot\space A=\frac{cos\space A}{sin\space A}\)

Therefore, \(tan\space cot\space A=\frac{1}{cot\space A}\)

Also, \(sec^2\space A=1+tan^2\space A\)

\(=1+\frac{1}{cot^2\space A}\)

\(=\frac{cot^2\space A+1}{1\space cot^2\space A}\)

\(sec\space A=\frac{\sqrt{cot^2\space A+1}}{cot\space A}\)

Question 2: Write all the other trigonometric ratios of \(\angle A\) in terms of \(sec\space A\).

Answer: 

We know that,

\(cos\space A=\frac{1}{sec\space A}\)

Also, \(sin^2\space A+cos^2\space A=1\)

\(sin^2 A=1-cos^2\space A\)

\(sin\space A=\sqrt{1-(\frac{1}{sec\space A})^2}\)

\(=\sqrt{\frac{sec\space A-1}{sec^2\space A}}=\frac{\sqrt{sec^2\space A-1}}{sec\space A}\)

\(tan^2\space A+1=sec^2\space A\)

\(tan^2\space A=sec^2\space A-1\)

\(tan\space A=\sqrt{sec^2\space A-1}\)

\(cot\space A=\frac{cos\space A}{sin\space A}=\frac{1}{\sqrt{sec^2\space A-1}}\)

\(=\frac{1}{\sqrt{sec^2\space A-1}}\)

\(cosec\space A=\frac{1}{sin\space A}=\frac{sec\space A}{\sqrt{sec^2\space A-1}}\)

Question 3:

Evaluate

(i) \(\frac{sin^2\space63^0+sin^2\space27^0}{cos^2\space17^0+cos^2\space73^0}\)

(ii) \(sin\space25^0\space cos\space65^0 + cos\space25^0\space sin\space65^0\)

(i) \(\frac{sin^2\space 63^0+sin^2\space 27^0}{cos^2\space17^0+cos^0\space 73^0}\)

\(=\frac{[sin(90^0-27^0)]^2+sin^2\space27^0}{[cos(90^0-73^0)]^2+cos^2\space73^0}\)

\(=\frac{[cos\space27^0]+sin^2\space27^0}{sin\space73^0+cos^2\space73^0}\)

\(=\frac{[cos\space27^0+sin^2\space27^0}{sin^2\space73^0+cos^2\space73^0}\)

\(=\frac{1}{1}\) (As\(sin^2\space A+cos^2\space A=1\))

\(=1\)

(ii) \(sin\space25^0\space cos\space65^0+cos\space25^0\space sin\space65^0\)

\(=(sin\space25^0)\lbrace cos(90^0-25^0)\rbrace+(cos\space25^0 \lbrace sin(90^0-25^0)\rbrace\)
\(=(sin\space25^0)(sin\space25^0)+(cos\space25^0)(cos\space25^0)\)
\(=sin^2\space25^0+cos^2\space25^0\)
\(=1\) (As \(sin^2\space A+cos^2\space A=1)\)

Question 4: Choose the correct option. Justify your choice.

(i) \(9\space sec^2\space A-9\space tan^2\space A=?\)

(a) \(1\)

(b) \(9\)

(c) \(8\)

(d) \(0\)

Answer: 

(i) \(9\space sec^2\space A-9\space tan^2\space A\)

\(=9(sec^2\space A-tan^2\space A\)

\(=9(1)\) [As \(sec^2\space A-tan^2\space A=1\)]

\(=9\)

(ii) \((1+tan\space\theta+sec\space\theta)(1+cot\space\theta-cosec\space\theta)\)

(i) \(0\)

(ii) \(1\)

(iii) \(2\)

(iv) \(-1\)

Answer: 

(ii) \((1+tan\space\theta+sec\space\theta)(1+cot\space\theta-cosec\space\theta)\)

\(=(1+\frac{sin\space\theta}{cos\space\theta}+\frac{1}{cos\space\theta})(1+\frac{cos\space\theta}{sin\space\theta}-\frac{1}{sin\space\theta})\)

\(=(\frac{cos\space\theta+sin\space\theta+1}{cos\space\theta})(\frac{sin\space\theta+cos\space\theta-1}{sin\space\theta})\)

\(=\frac{(sin^2\space\theta+cos^2\space\theta+2\space2\space sin\space\theta\space cos\space\theta-1}{sin\space\theta\space cos\space\theta}\)

\(=\frac{1+2\space\theta\space cos\space\theta-1}{sin\space\theta\space cos\space\theta}\)

\(=\frac{2\space\theta\space cos\space\theta}{sin\space\theta cos\space\theta}=2\)

(iii) \((sec\space A+tan\space A)(1-sin\space A)=?\)

(a) \(sec\space A\)

(b) \(sin\space A\)

(c) \(cosec\space A\)

(d) \(cos\space A\)

Answer: 

\((sec\space A+tan\space A)(1-sin\space A)\)

\(=(\frac{1}{cos\space A}+\frac{sin\space A}{cos\space A})(1-sin\space A)\)

\(=(\frac{1+sin\space A}{(cos\space A)(1-sin\space A)})\)

\(=\frac{1-sin^2\space A}{cos\space A}=\frac{cos^2\space A}{cos\space A}\)

\(=cos\space A\)

(iv) \(\frac{1+tan^2\space A}{1+cot^2\space A}\)

(a) \(sec^2\space A\)

(b) \(-1\)

(c) \(cot^2\space A\)

(d) \(tan^2\space A\)

Answer: 

\(\frac{1=tan^2\space A}{1+cot^2\space A}=\frac{1+\frac{sin^2\space A}{cos^2\space A}}{1+\frac{cos^2\space A}{sin^2\space A}}\)
\(=\frac{\frac{cos^2\space A+sin^2\space A}{cos^2\space A}}{\frac{sin^2\space A+cos^2\space A}{sin^2\space A}}=\frac{\frac{1}{cos^2\space A}}{\frac{1}{sin^2\space A}}\)
\(=\frac{sin^2\space A}{cos^2\space A}=tan^2\space A\)

Question 5: Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Answer:

(i) \((cosec\space\theta-cot\space\theta)^2=\frac{1+cos\space\theta}{1+cos\space\theta}\)
\(L.H.S.=(cosec\space\theta-cot\space\theta)^2\)
\(=(\frac{1}{sin\space\theta}-\frac{cos\space\theta}{sin\space\theta})^2\)
\(=\frac{(1-cos\space\theta)^2}{1-cos^2\space\theta}=\frac{(1-cos\space\theta)^2}{(1-cos\space\theta)(1+cos\space\theta)}=\frac{1-cos\space\theta}{1+cos\space\theta}\)
\(=R.H.S\)

(ii) \(\frac{cos\space A}{1+sin\space A}+\frac{1+sin\space A}{cos\space A}=2\space sec\space A\)
\(L.H.S=\frac{cos\space A}{1+sin\space A}+\frac{1+sin\space A}{cos\space A}\)
\(=\frac{cos^2\space A+(1-sin\space A)^2}{(1+sin\space A)(cos\space A)}\)
\(=\frac{sin^2\space A+cos^2\space A+1+2\space sin\space A}{(1+sin\space A)(cos\space A)}\)
\(=\frac{1+1+2\space sin\space A}{(1+sin\space A)(cos\space A)}=\frac{2+2\space sin\space A}{(1+sin\space A)(cos\space A)}\)
\(=\frac{2(1+sin\space A)}{(1+sin\space A)(cos\space A)}=\frac{2}{cos\space A}=2\space sec\space A\)
\(=R.H.S\)

(iii) \(\frac{tan\space\theta}{1-cot\space\theta}+\frac{cot\space\theta}{1-tan\space\theta}=1+sec\space\theta\space cosec\space\theta\)
\(L.H.S=\frac{tan\space\theta}{1-cot\space\theta}+\frac{cot\space\theta}{1-tan\space\theta}\)
\(=\frac{\frac{sin\space\theta}{cos\space\theta}}{1-\frac{cos\space\theta}{sin\space\theta}}+\frac{\frac{cos\space\theta}{sin\space\theta}}{1-\frac{sin\space\theta}{cos\space\theta}}\)
\(=\frac{\frac{sin\space\theta}{cos\space\theta}}{\frac{sin\space\theta-cos\space\theta}{sin\space\theta}}+\frac{\frac{cos\space\theta}{sin\space\theta}}{\frac{cos\space\theta-sin\space\theta}{cos\space\theta}}\)
\(=\frac{sin^2\space\theta}{cos\space\theta(sin\space\theta-cos\space\theta)}+\frac{cos^2\space\theta}{sin\space\theta(sin\space\theta-cos\space\theta)}\)
\(=\frac{1}{(sin\space\theta-cos\space\theta)}[\frac{sin^2\space\theta}{cos\space\theta}-\frac{cos^2\space\theta}{sin\space\theta}]\)
\(=(\frac{1}{sin\space\theta-cos\space\theta})[\frac{sin^3\space\theta-cos^3\space\theta}{sin\space\theta\space cos\space\theta}]\)
\(=(\frac{1+sin\space\theta\space cos\space\theta}{sin\space\theta\space cos\space\theta})\)
\(=sec\space\theta\space cosec\space\theta\)
\(=R.H.S\)

(iv) \(\frac{1+sec\space A}{sec\space A}=\frac{sin^2\space A}{1-cos\space A}\)
\(L.H.S=\frac{1+sec\space A}{sec\space A}=\frac{1+\frac{1}{cos\space A}}{\frac{1}{cos\space A}}\)
\(=\frac{\frac{cos\space A+1}{cos\space A}}{\frac{1}{cos\space A}}=(cos\space A+1)\)
\(=\frac{(1-cos\space A)(1+cos\space A)}{\frac{1}{cos\space A}}\)
\(=\frac{(1-cos\space A)(1+cos\space A)}{(1-cos\space A)}\)
\(=\frac{1-cos^2\space A}{1-cos\space A}=\frac{sin^2\space A}{1-cos\space A}\)
\(=R.H.S\)

(v) \(\frac{cos\space A-sin\space A+1}{cos\space A+sin\space A-1}=cosec\space A+cot\space A\)
Using the identity \(cisec^2=1+cot^2\)
\(L.H.S=\frac{cos\space A-sin\space A+1}{cos\space A+sin\space A-1}\)
\(=\frac{\frac{cos\space A}{sin\space A}-\frac{sin\space A}{sin\space A}+\frac{1}{sin\space A}}{\frac{cos\space A}{sin\space A}+\frac{sin\space A}{sin\space A}+\frac{1}{sin\space A}}\)
\(=\frac{\lbrace(cot\space A)-(1-cosec\space A)\rbrace\lbrace(cot\space A)-(1-cosec\space A)\rbrace}{\lbrace(cot\space A)+91-cosec\space A)\rbrace\lbrace(cot\space A)-(1-cosec\space A)\rbrace}\)
\(=\frac{(cot\space A-1+cosec\space A)^2}{(cot\space A)^2-(1-cosec\space A)^2}\)
\(=\frac{cot^2\space A+1+cosec^2\space A-2\space cot\space A-2\space cosec\space A+2\space cot\space A\space cosec\space A}{cot^2\space A-(1+cosec^2\space A-2\space cosec\space A)}\)
\(=\frac{2\space cosec^2\space A+2\space cot\space A\space cosec\space A-2\space cot\space A-2\space cosec\space A}{cot^2\space A-1-cosec^2\space A+2\space cosec\space A}\)
\(=\frac{2\space cosec^2\space A+2\space cot\space A\space cosec A-2\space cot\space A-2\space cosec\space A}{1\space cot^2\space A-1-cosec^2\space A+2\space cosec\space A}\)
\(\frac{2\space cosec\space A(cosec\space A+cot\space A)-2(cot\space A+cosec\space A)}{cot^2\space A \space cosec^2\space A-1+2\space cosec\space A}\)
\(=\frac{(cosec\space A+cot\space A)(2\space cosec\space A-2)}{-1+2\space cosec\space A}\)
\(=\frac{(cosec\space A+cot\space A)(2\space cosec\space A-2)}{(2\space cosec\space A-2)}\)
\(=cosec\space A=cot\space A\)
\(R.H.S\)

(vi) \(\sqrt{\frac{1+sin\space A}{1-sin\space}}=sec\space A+tan\space A\)
\(L.H.S=\sqrt{\frac{1}{2}}\)
\(L.H.S=\sqrt{\frac{1+sin\space A(1+sin\space A)}{1-sin\space A(1+sin\space A}}\)
\(=\frac{(1+sin\space A)}{\sqrt{1-sin^2\space A}}=\frac{1+sin\space A}{\sqrt{cos^2\space A}}\)
\(=\frac{1+sin\space A}{cos\space A}=sec\space A+tan\space A\)
\(=R.H.S\)
(vii) \(\frac{sin\space\theta-2\space sin^2\space\theta}{2\space cose^3\theta-cos\space\theta}\)
\(L.H.S=\frac{sin\space\theta-2\space\sin^3\space\theta}{2\space cos^3\theta-cos\space\theta}\)
\(=\frac{sin\space\theta(1-2\space sin^3\space\theta)}{cos\space\theta(2\space cos^2\space\theta-1)}\)
\(=\frac{sin\space\theta\times(1-2\space sin^2\space\theta)}{cos\space\theta\times \lbrace2(1-sin^2\space\theta)-1\rbrace}\)
\(=\frac{sin\space\theta\times(1-2\space sin^2\space\theta}{cos\space\theta\times(1\times2\space sin^2\space \theta)}\)
\(=tan\space\theta=R.H.S\)

(vii) \((sin\space A+cosec\space A)62+(cos\space A+sec\space A)^2=7=tan^2\space A+cot^2\space A\)
\(L.H.S=(cosec\space A-sin\space A)(sec\space A-cos\space A)\)
\(=(\frac{1}{sin\space A}-sin\space A)(\frac{1}{cos\space A}-cos\space A)\)
\(=(\frac{1-sin^2\space A}{sin\space A})(\frac{1-cos^2\space A}{cos\space A})\)
\(=\frac{(cos^2\space A)(sin^2\space A)}{sin\space A\space cos\space A}\)
\(=sin\space A\space cos\space A\)
\(R.H.S.=\frac{1}{tan\space A+cot\space A}\)
\(=\frac{1}{\frac{sin\space A}{cos\space A}}+\frac{cos\space A}{sin\space A}=\frac{1}{\frac{sin^2\space A+cos^2\space A}{sin\space A\space cos\space A}}\)
\(=\frac{sin\space A\space cose\space A}{sin^2\space A+cos^2\space A}=sin\space A\space cose\space A\)
Hence, \(L.H.S=R.H.S\)

(x) \((\frac{1+tan^2\space A}{1+cot^2\space A})=(\frac{1-tan\space A}{1-cot\space A})^2=tan^2\space A\)
\(\frac{1+tan^2\space A}{1+cot^2\space A}=\frac{1+tan^2\space A-2\space tan\space A}{1+cot^2\space A-2\space cot \space A}\)
\(=\frac{sec^2\space A-2\space tan\space A}{cosec^2\space A-2\space cot \space A}\)
\(=\frac{\frac{1}{cos^2\space A}-\frac{2\space sin\space A}{cos\space A}}{\frac{1}{1\space sin^2\space A}-\frac{2\space cos\space A}{sin\space A}}=\frac{\frac{1-2\space sin\space A\space cos\space A}{cos^2\space A}}{\frac{1-2\space sin\space A\space cos\space A}{sin^2\space A}}\)
\(=\frac{sin^2\space A}{cos^2\space A}=tan^2\space A\)