Practicing ML Aggarwal Class 11 Solutions ISC Chapter 3 Trigonometry Ex 3.2 is the ultimate need for students who intend to score good marks in examinations.
ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.2
Short and long answer questions (1 to 25) :
Prove the following (1 to 17) trigonometrical identities :
Question 1.
(i) sin x cot x + sin x cosec x = 1 + cos x
(ii) \(\frac{1}{1+\tan ^2 x}+\frac{1}{1+\cot ^2 x}\) = 1.
Solution:
(i) L.H.S. = sin x cot x + sin x cosec x
= sin x \(\frac{\cos x}{\sin x}\) + \(\frac{\sin x}{\sin x}\)
= cos x + 1
= R.H.S.
(ii) L.H.S. = \(\frac{1}{1+\tan ^2 x}+\frac{1}{1+\cot ^2 x}\)
= \(\frac{1}{1+\tan ^2 x}+\frac{1}{1+\frac{1}{\tan ^2 x}}\)
[∵ cot x = \(\frac{1}{\tan x}\)]
= \(\frac{1}{1+\tan ^2 x}+\frac{\tan ^2 x}{1+\tan ^2 x}\)
= \(\frac{1+\tan ^2 x}{1+\tan ^2 x}\)
= 1
= R.H.S.
Question 2.
(i) cot2 x – \(\frac{1}{\sin ^2 x}\) + 1 = 0
(ii) sec x (1 – sin x) (sec x + tan x) = 1
Solution:
(i) L.H.S. = cot2 x – \(\frac{1}{\sin ^2 x}\) + 1
= cot2 x – cosec2 x + 1
= (1 + cot2 x) – cosec2 x
= cosec2 x – cosec2 x
= R.H.S.
(ii) LH.S. = sec x (1 – sin x) (sec x + tan x)
= \(\frac{1}{\cos x}(1-\sin x)\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)\)
= \(\frac{1}{\cos x} \frac{(1-\sin x)(1+\sin x)}{\cos x}\)
= \(\frac{1-\sin ^2 x}{\cos ^2 x}\)
= \(\frac{\cos ^2 x}{\cos ^2 x}\)
= 1
= R.H.S.
Question 3.
(i) \(\)
(ii) \(\)
Solution:
(i) L.H.S. = \(\frac{1+\sec x}{\sec x}\)
= \(\frac{1+\frac{1}{\cos x}}{\frac{1}{\cos x}}\)
= (1 + cos x) \(\frac{1-\cos x}{1-\cos x}\)
= \(\frac{1-\cos ^2 x}{1-\cos x}\)
= \(\frac{\sin ^2 x}{1-\cos x}\)
= R.H.S.
(ii) L.H.S. = \(\frac{\tan x+\sin x}{\tan x-\sin x}\)
= \(\frac{\sin x \sec x+\sin x}{\sin x \sec x-\sin x}\)
= \(\frac{\sec x+1}{\sec x-1}\)
= R.H.S.
Question 4.
(i) \(\frac{\sec x-1}{\sec x+1}=\frac{1-\cos x}{1+\cos x}\)
(ii) sec2 x + cosec2 x = sec2 x cosec2 x.
Solution:
(i) L.H.S. = \(\frac{\sec x-1}{\sec x+1}\)
= \(\frac{\frac{1}{\cos x}-1}{\frac{1}{\cos x}+1}\)
= \(\frac{\frac{1-\cos x}{\cos x}}{\frac{1+\cos x}{\cos x}}\)
= \(\frac{1-\cos x}{1+\cos x}\)
= R.H.S.
(ii) L.H.S. = sec2 x + cosec2 x
= \(\frac{1}{\cos ^2 x}+\frac{1}{\sin ^2 x}\)
= \(\frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x}\)
= \(\frac{1}{\sin ^2 x \cos ^2 x}\)
= sec2 x + cosec2 x
= R.H.S.
Question 5.
(i) tan2 x – sin2 x = tan2 x sin2 x
(ii) \(\frac{\cos x}{1-\tan x}-\frac{\sin ^2 x}{\cos x-\sin x}\) = cos x + sin x
Solution:
(I) L.H.S. = tan2 x – sin2 x
= sin2 x [\(\frac{1}{\cos ^2 x}\) – 1]
= sin2 x \(\frac{\left(1-\cos ^2 x\right)}{\cos ^2 x}\)
= sin2 x \(\frac{\sin ^2 x}{\cos ^2 x}\)
= sin2 x tan2 x
= R.H.S.
(ii) L.H.S. = \(\frac{\cos x}{1-\tan x}-\frac{\sin ^2 x}{\cos x-\sin x}\)
= \(\frac{\cos ^2 x}{\cos x-\sin x}-\frac{\sin ^2 x}{\cos x-\sin x}\)
= \(\frac{\cos ^2 x-\sin ^2 x}{\cos x-\sin x}\)
= \(\frac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x-\sin x}\)
= cos x + sin x
= R.H.S.
Question 6.
(i) \(\sqrt{\frac{1+\cos x}{1-\cos x}}\) = cosec x + cot x
(ii) \(\frac{\sin x \tan x}{1-\cos x}\) = 1 + sec x
Solution:
(i) L.H.S. = \(\sqrt{\frac{1+\cos x}{1-\cos x}}\)
= \(\sqrt{\frac{(1+\cos x)(1+\cos x)}{(1-\cos x)(1+\cos x)}}\)
= \(\frac{1+\cos x}{\sqrt{1-\cos ^2 x}}\)
= \(\frac{1+\cos x}{\sin x}\)
= \(\frac{1}{\sin x}+\frac{\cos x}{\sin x}\)
= cosec x + cot x
= R.H.S
(ii) L.H.S. = \(\frac{\sin x \tan x}{1-\cos x}\)
= \(\frac{\sin ^2 x}{\cos x(1-\cos x)}\)
= \(\frac{1-\cos ^2 x}{\cos x(1-\cos x)}\)
= \(\frac{(1-\cos x)(1+\cos x)}{\cos x(1-\cos x)}\)
= \(\frac{1+\cos x}{\cos x}\)
= \(\frac{1}{\cos x}\) + 1
= sec x + 1
= R.H.S.
Question 7.
(i) \(\frac{1+\cos x}{1-\cos x}=\frac{\tan ^2 x}{(\sec x-1)^2}\)
(ii) \(\frac{\tan ^2 x}{\tan ^2 x-1}+\frac{\ {cosec}^2 x}{\sec ^2 x-\ {cosec}^2 x}\) = \(\frac{1}{\sin ^2 x-\cos ^2 x}\)
Solution:
(i) R.H.S.
(ii)
Question 8.
(i) sin4 x – cos4 x = 1 – 2 cos2 x
(ii) sec4 x – tan4 x = 1 + 2 tan2 x
(iii) sin6 x + cos6 x = 1 – 3 sin2 x cos2 x.
Solution:
(i) L.H.S. = sin4 x – cos4 x
= (sin2 x – cos2 x) (sin2 x + cos2 x)
= sin2 x – cos2 x
= 1 – cos2 x – cos2 x
= 1 – 2 cos2 x
= R.H.S.
(ii) L.H.S. = sec4 x – tan4 x
= (sec2 x – tan2 x) (sec2 x + tan2 x)
= 1 . (sec2 x + tan2 x)
[∵ sec2 θ = 1 + tan2 θ]
= 1 + 2 tan2 x
= R.H.S.
(iii) L.H.S. = sin6 x + cos6 x
= (sin2 x)3 + (cos2 x)3
= (sin x + cos x)3 – 3 sin x cos x [sin x + cos x]
[∵ a3 + b3 = (a + b)3 – 3ab (a + b)]
= 1 -3 sin2 x cos2 x
= R.H.S.
[∵ sin2 θ + cos2 θ = 1]
Question 9.
(i) \(\frac{1+\tan x}{\sin x}+\frac{1+\cot x}{\cos x}\) = 2 (sec x + cosec x)
(ii) (sin x + cosec x)2 + (cos x + sec x)2 = tan2 x + cot2 x + 7
Solution:
(i) L.H.S. = \(\frac{1+\tan x}{\sin x}+\frac{1+\cot x}{\cos x}\)
= \(\frac{1+\tan x}{\sin x}+\frac{1+\frac{1}{\tan x}}{\cos x}\)
= \(\frac{1+\tan x}{\sin x}+\frac{1+\tan x}{\tan x \cdot \cos x}\)
= \(2\left[\frac{1+\tan x}{\sin x}\right]\)
= 2 \(\left[\frac{1}{\sin x}+\frac{\tan x}{\sin x}\right]\)
= 2 [cosec x + sec x]
= R.H.S.
(ii) L.H.S. = (sin x + cosec x)2 + (cos x + sec x)2
= sin2 x + cosec2 x + 2 sin x cosec x + cos2 x + sec2 x + 2 cos x sec x
= (sin2 x + cos2 x) + (1 + cot2 x) + 2 + (1 + tan2 x) + 2
= 7 + tan2 x + cot2 x
= R.H.S.
Question 10.
(i) cosec6 x – cot6 x = 3 cot2 x cosec2 x + 1
(ii) sec6 x – tan6 x = 1 + 3 tan2 x + 3 tan4 x.
Solution:
(i) L.H.S. cosec6 x – cot6 x
= (cosec2 x)3 – (cot2 x)3
= (cosec2 x – cot2 x)3 + 3 cosec2 x cot2 x (cosec2 x – cot2 x)
[∵ a3 – b3 = (a – b)3 + 3 ab (a – b)]
= 13 + 3 cosec2 x cot2 x . 1
[∵ cosec2 θ = 1 + cot2 θ]
= 1 + 3 cosec2 x cot2 x
= R.H.S.
(ii) L.H.S. = sec6 x – tan6 x
= (sec2 x)3 – (tan2 x)3
= (sec2 x – tan2 x)3 + 3 sec2 x tan2 x (sec2 x – tan2 x)
[∵ a3 – b3 = (a – b)3 + 3ab (a – b)]
= 1 + 3 tan2 x (1 + tan2 x)
[∵ sec2 x = 1 + tan2 x]
= 1 + 3 tan2 x + 3 tan4 x
= R.H.S.
Question 11.
(i) \(\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}\) = 1 + tan x + cot x = 1 + sec x cosec x
(ii) \(\frac{1+\cos x+\sin x}{1+\cos x-\sin x}=\frac{1+\sin x}{\cos x}\)
Solution:
(i)
(ii)
Question 12.
(i) \(\frac{\cos ^2 x}{1-\tan x}+\frac{\sin ^3 x}{\sin x-\cos x}\) = 1 + sin x cos x
(ii) cot2 x . \(\frac{\sec x-1}{1+\sin x}\) + sec2 x . \(\frac{\sin x-1}{1+\sec x}\) = 0
Solution:
(i) L.H.S. = \(\frac{\cos ^2 x}{1-\tan x}+\frac{\sin ^3 x}{\sin x-\cos x}\)
= \(\frac{\cos ^3 x}{\cos x-\sin x}+\frac{\sin ^3 x}{\sin x-\cos x}\)
= \(\frac{\cos ^3 x-\sin ^3 x}{\cos x-\sin x}\)
= \(\frac{(\cos x-\sin x)\left(\cos ^2 x+\sin ^2 x+\cos x \sin x\right)}{\cos x-\sin x}\)
= 1 + sin x cos x
= R.H.S.
(ii)
Question 13.
(i) \(\frac{\cot x+\ {cosec} x-1}{\cot x-\ {cosec} x+1}=\frac{1+\cos x}{\sin x}\)
(ii) \(\frac{\sin x}{\cot x+\ {cosec} x}=2+\frac{\sin x}{\cot x-\ {cosec} x}\)
Solution:
(i)
(ii)
Question 14.
(i) (sin x + cos x) (sec x + cosec x) = 2 + sec x cosec x
(ii) (cosec x – sin x) (sec x – cos x) (tan x + cot x) = 1.
Solution:
(i) L.H.S. = (sin x + cos x) (sec x + cosec x)
= (sin x + cos x) \(\left(\frac{1}{\cos x}+\frac{1}{\sin x}\right)\)
= \(\frac{(\sin x+\cos x)^2}{\sin x \cos x}\)
= \(\frac{\sin ^2 x+\cos ^2 x+2 \sin x \cos x}{\sin x \cos x}\)
= \(\frac{1+2 \sin x \cos x}{\sin x \cos x}\)
= \(\frac{1}{\sin x \cos x}\) + 2
= sec x cosec x + 2
= R.H.S.
(ii) L.H.S. = (cosec x – sin x) (sec x – cos x) (tan x + cot x)
= \(\left(\frac{1}{\sin x}-\sin x\right)\left(\frac{1}{\cos x}-\cos x\right)\left[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right]\)
= \(\frac{\left(1-\sin ^2 x\right)}{\sin x}\left(\frac{1-\cos ^2 x}{\cos x}\right)\left(\frac{\sin ^2 x+\cos ^2 x}{\sin x \cos x}\right)\)
= \(\left(\frac{\cos ^2 x}{\sin x}\right)\left(\frac{\sin ^2 x}{\cos x}\right)\left(\frac{1}{\sin x \cos x}\right)\)
= 1
= R.H.S.
Question 15.
(i) \(\frac{\sin ^3 x+\cos ^3 x}{\sin x+\cos x}+\frac{\sin ^3 x-\cos ^3 x}{\sin x-\cos x}\) = 2
(ii) \(\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}\) = 1 + sec x cosec x
Solution:
(i) L.H.S. = \(\frac{\sin ^3 x+\cos ^3 x}{\sin x+\cos x}+\frac{\sin ^3 x-\cos ^3 x}{\sin x-\cos x}\)
= \(\frac{(\sin x+\cos x)\left(\sin ^2 x-\sin x \cos x+\cos ^2 x\right)}{\sin x+\cos x}\) + \(\frac{(\sin x-\cos x)\left(\sin ^2 x+\sin x \cos x+\cos ^2 x\right)}{\sin x-\cos x}\)
[∵ a3 ± b3 = (a ± b) (a2 ± ab + b2)]
= sin2 x + cos2 x – sin x cos x + sin2 x + cos2 x + sin x cos x
= 2 (sin2 x + cos2 x)
= 2 × 1 = 2
(ii)
Question 16.
(i) \(\frac{\tan x+\tan y}{\cot x+\cot y}\) = tan x tan y
(ii) \(\frac{\cot x+\tan y}{\cot y+\tan x}\) = cot x tan y
Solution:
(i) L.H.S. = \(\frac{\tan x+\tan y}{\cot x+\cot y}\)
= \(\frac{\tan x+\tan y}{\frac{1}{\tan x}+\frac{1}{\tan y}}\)
= \(\frac{(\tan x+\tan y)}{\frac{\tan y+\tan x}{\tan x \tan y}}\)
= tan x tan y
= R.H.S.
(ii) L.H.S. = \(\frac{\cot x+\tan y}{\cot y+\tan x}\)
= \(\frac{\cot x+\tan y}{\frac{1}{\tan y}+\frac{1}{\cot x}}\)
= \(\frac{\cot x+\tan y}{\frac{\cot x+\tan y}{\cot x \tan y}}\)
= cot x tan y
= R.H.S.
Question 17.
(i) sin2 x cos2 y – cos2 x sin2 y = sin2 x – sin2 y
(ii) tan2 x sec2 y – sec2 x tan2 y = tan2 x – tan2 y.
Solution:
(j) L.H.S. = sin2 x cos2 y – cos2 x sin2 y
= sin2 x (1 – sin2 y) – (1 – sin2 x) sin2 y
= sin2 x – sin2 x sin2 y – sin2 y + sin2 x sin2 y
= sin2 x – sin2 y
= R.H.S.
(ii) L.H.S. = tan2 x sec2 y – sec2 x tan2 y
= tan2 x (1 + tan2 y) – (1 + tan2 x) tan2 y
= tan2 x + tan2 x tan2 y – tan2 y – tan2 x tan2 y
= tan2 x – tan2 y
= R.H.S.
Question 18.
If cos x + cos2 x = 1, prove that sin2 x + sin4 x= 1.
Solution:
Given cos x + cos2 x = 1
⇒ cos x = 1 – cos2 x
⇒ cos x = sin2 x
On squaring both sides ; we have
cos2 x = sin4 x
⇒ 1 – sin2 x = sin4 x
⇒ sin4 x + sin2 x = 1.
Eliminate t between the following (19 to 23) equations :
Question 19.
x = a sec t, y = b tan t.
Solution:
Given x = a sec t
⇒ \(\frac{x}{a}\) = sec t …………………(1)
and y = b tan t
⇒ \(\frac{y}{b}\) = tan t ……………….(2)
On squaring (1) and (2) and subtracting ; we have
\(\left(\frac{x}{a}\right)^2-\left(\frac{y}{b}\right)^2\) = sec2 t tan2 t
⇒ \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1.
Question 20.
x = h + a cos t, y = k + sin t
Solution:
Given x = h + a cos t
⇒ \(\frac{x-h}{a}\) = cos t ………………..(1)
and y = k + sin t
⇒ y – k = sin t ………………….(2)
On squaring (1) and (2) and then adding
\(\left(\frac{x-h}{a}\right)^2\) + (y – k)2 = cos2 t + sin2 t
⇒ \(\left(\frac{x-h}{a}\right)^2\) + (y – k)2 = 1
Question 21.
sec t + tan t = m, sec t – tan t = n.
Solution:
Given sec t + tan r = m ……………………(1)
and sec t – tan t = n ………………..(2)
On adding (1) and (2) ; we have
2 sec t = m + n
⇒ sec t = \(\frac{m+n}{2}\) ……………..(3)
eqn. (1) – eqn. (2) gives ;
2 tan t = m – n
⇒ tan t = \(\frac{m-n}{2}\) ………………..(4)
Squaring eqn. (3) and (4) and subtracting ; we have
sec2 t – tan2 t = \(\left(\frac{m+n}{2}\right)^2-\left(\frac{m-n}{2}\right)^2\)
(m + n)2 – (m – n)2 = 4
⇒ 2mn + 2mn = 4
⇒ mn = 1
Question 22.
a cosec t + b cot t = x, a cot t + b cosec t = y.
Solution:
Given a cosec t + b cot t – x = 0 ………………..(1)
and b cosec t + a cot t – y = 0 ………………..(2)
using cross multiplication method, we have
\(\frac{\ {cosec} t}{a x-b y}=\frac{\cot x}{a y-b x}\)
= \(\frac{1}{a^2-b^2}\)
and cosec = \(\frac{a x-b y}{a^2-b^2}\) ……………….(3)
and cot t = \(\frac{a y-b x}{a^2-b^2}\) ………………..(4)
On squaring eqn. (3) and eqn. (4) and subtracting ; we have
cosec2 t – cot2 t = \(\left(\frac{a x-b y}{a^2-b^2}\right)^2-\left(\frac{a y-b x}{a^2-b^2}\right)^2\)
⇒ (a2 – b2)2 = a2 (x2 – y2) + b2 (y2 – x2)
⇒ a2 – b2 = x2 – y2
Question 23.
cot t + cos t = m, cot t – cos t = n.
Solution:
Given cot t + cos t = m ………………(1)
and cot t – cos t = n …………………(2)
On adding (1) and (2) ; we have
2 cot t = m + n
⇒ tan t = \(\frac{2}{m+n}\) …………………(3)
Subtracting eqn. (2) from (1) ; we have
2 cos t = m — n
⇒ sec t = \(\frac{2}{m-n}\) …………………(4)
We know that, sec2 t = 1 + tan2 t
⇒ \(\left(\frac{2}{m-n}\right)^2=1+\left(\frac{2}{m+n}\right)^2\)
⇒ \(\frac{-4}{(m-n)^2}-\frac{4}{(m+n)^2}\) = 1
⇒ \(\frac{4\left[(m+n)^2-(m-n)^2\right]}{\left(m^2-n^2\right)^2}\) = 1
⇒ 4 (4 mn) = (m2 – n2)2
⇒ (m2 – n2)2 = 16mn.
Question 24.
If x = a sin t and y = b tan t, prove that \(\frac{a^2}{x^2}-\frac{b^2}{y^2}\) = 1.
Solution:
Given x = a sin t
⇒ \(\frac{x}{a}\) = sin t ……………….(1)
⇒ \(\frac{a}{x}\) = cosec t
y = b tan t
⇒ \(\frac{y}{b}\) = tan t ……………….(2)
⇒ \(\frac{b}{y}\) = cot t
Squaring (1) and (2) and then subtracting the resulting eqn’s ;
⇒ \(\left(\frac{a}{x}\right)^2-\left(\frac{b}{y}\right)^2\) = cosec2 t – cot2 t
⇒ \(\frac{a^2}{x^2}-\frac{b^2}{y^2}\) = 1
Question 25.
If m = \(\frac{\cos \alpha}{\cos \beta}\) and n = \(\frac{\cos \alpha}{\sin \beta}\), show that (m2 + n2) cos2 β = n2.
Solution:
Given m = \(\frac{\cos \alpha}{\cos \beta}\)
⇒ cos α = m cos β ………………………(1)
and n = \(\)\frac{\cos \alpha}{\sin \beta}[/latex
⇒ cos α = n sin β ……………………..(2)
From (1) and (2) ; we have
m cos β = n sin β
On squaring ; we have
m2 cos2 β = n2 sin2 β
= n2 (1 – cos2 β)
⇒ (m2 + n2) cos2 β = n2