Question

Prove that ko secant squared theta + secant squared theta equal to cos square theta second squared theta

Answer 1

Answer:

{sec}^{2} \theta + {cosec}^{2} \theta = {sec}^{2} \theta. {cosec}^{2} \thetasec

2

θ+cosec

2

θ=sec

2

θ.cosec

2

θ

L.H.S.

\begin{gathered}= \frac{1}{ {cos}^{2} \theta} + \frac{1}{ {sin}^{2} \theta} \\ = \frac{ {sin}^{2} \theta \: + {cos}^{2} \theta}{ {cos}^{2} \theta. {sin}^{2} \theta } \\ = \frac{1}{ {cos}^{2} \theta. {sin}^{2} \theta } \\ = {sec}^{2} \theta. {cosec}^{2} \theta\end{gathered}

=

cos

2

θ

1

+

sin

2

θ

1

=

cos

2

θ.sin

2

θ

sin

2

θ+cos

2

θ

=

cos

2

θ.sin

2

θ

1

=sec

2

θ.cosec

2

θ

= R.H.S.

Hence, proved