Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • If sinθ – cosθ = 0, then the value of (sin power 4 θ + cos poer 4θ) is (A) 1 (B) 3 by 4 (C) 1 by 2 (D) 1 by 4

If sinθ – cosθ = 0, then the value of (sin4θ + cos4θ) is

(A) 1                (B)3/4                     (C) 1/2                   (D) 1/4

Answers (1)

\sin\theta-\cos\theta=0
squaring both sides we get
(\sin\theta-\cos\theta)^2=0
\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta=0
(\therefore (a – b)2 = a2 + b2 – 2ab)
\sin^2\theta+\cos^2\theta=2\sin\theta\cos\theta.......(1)
1=2\sin\theta\cos\theta\because (\sin^2\theta+\cos^2\theta=1)
\frac{1}{2}=\sin \theta \cos \theta
Squaring both sides we get

\frac{1}{4}=\sin^{2} \theta \cos^{2} \theta       …(2)
Now squaring both side of equation (1) we get
\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )^{2}= \left ( 2\sin \theta \cos \theta \right )^{2}
\left ( \sin ^{2}\theta \right )^{2}+\left ( \cos ^{2}\theta \right )^{2}+2\sin ^{2}\theta \cdot \cos ^{2}\theta = 4\sin ^{2}\theta \cos ^{2}\theta
\left [ \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right ]
\left ( \sin ^{4}\theta \right )+\left ( \cos ^{4}\theta \right )= 4\sin ^{2}\theta \cos ^{2}\theta -2\sin ^{2}\theta \cos ^{2}\theta
\sin ^{4}\theta +\cos ^{4}\theta = 2\sin ^{2}\theta \cos ^{2}\theta
  (Use equation (2))
\sin ^{4}\theta +\cos ^{4}\theta = 2\left ( \frac{1}{4} \right )
\sin ^{4}\theta +\cos ^{4}\theta =\frac{1}{2}
Hence option (C) is correct.

Posted by

infoexpert27

View full answer

Similar Questions

Latest Question