2+tan²theta + 1/ tan² theta = 1/sin²theta - sin^4 theta
Answers
Answer:
Solution:
Given
LHS=\begin{gathered}\left(1+\frac{1}{tan^{2}\theta}\right)\left(1+\frac{1}{cot^{2}\theta}\right)\\ < /p > < p > = \left(1+cot^{2}\theta\right)\cdot\left(1+tan^{2}\theta\right)\end{gathered}
(1+
tan
2
θ
1
)(1+
cot
2
θ
1
)
</p><p>=(1+cot
2
θ)⋅(1+tan
2
θ)
_________________________
We know,
i) \frac{1}{tan\theta}=cot\theta
tanθ
1
=cotθ
ii)\frac{1}{cot\theta}=tan\theta
cotθ
1
=tanθ
Trigonometric identities :
1) 1+\cot^{2}\theta = cosec^{2}\theta1+cot
2
θ=cosec
2
θ
2) 1+tan^{2}\theta = sec^{2}\theta1+tan
2
θ=sec
2
θ
3) cos^{2}\theta=1-sin^{2}\thetacos
2
θ=1−sin
2
θ
and
a) cosec\theta = \frac{1}{sin\theta}cosecθ=
sinθ
1
b) sec\theta = \frac{1}{cos\theta}secθ=
cosθ
1
_________________________
=cosec^{2}\theta sec^{2}\thetacosec
2
θsec
2
θ
= \frac{1}{sin^{2}\theta}\cdot\frac{1}{cos^{2}\theta}
sin
2
θ
1
⋅
cos
2
θ
1
= \frac{1}{sin^{2}\theta\left(1-sin^{2}\theta\right)}
sin
2
θ(1−sin
2
θ)
1
= \frac{1}{sin^{2}\theta-sin^{4}\theta}
sin
2
θ−sin
4
θ
1
=RHSRHS
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