Question

Prove that Cosec²α+Sec²α=Cosec²α×Sec²α​

Answer 1

Answer:

To prove that,

${ \csc }^{2} \alpha + { \sec }^{2} \alpha = { \csc }^{2} \alpha \times { \sec }^{2} \alpha$

Let's start from Left Hand Side.

Therefore, we will get,

L.H.S =

${ \csc }^{2} \alpha + { \sec}^{2} \alpha \\ \\ = \dfrac{1}{ { \sin }^{2} \alpha } + \dfrac{1}{ { \cos }^{2} \alpha } \\ \\ = \frac{ { \cos }^{2} \alpha + { \sin }^{2} \alpha }{ { \sin }^{2} \alpha \times { \cos }^{2} \alpha } \\ \\ = \frac{1}{ { \sin }^{2} \alpha \times { \cos }^{2} \alpha } \\ \\ = { \csc}^{2} \alpha \times { \sec }^{2} \alpha$

= R.H.S

Thus, L.H.S = R.H.S

Hence, Proved.

Key Concepts :-

• $\csc \alpha = \frac{1}{ \sin( \alpha ) }$
• $\sec( \alpha ) = \frac{1}{ \cos( \alpha ) }$
• ${ \sin }^{2} \alpha + { \cos}^{2} \alpha = 1$