Question

$tan ^{2} \beta \times \csc {}^{2} \beta = \sec {}^{2} \beta$
prove the following.​

Answer 1

$\large{\boxed{\text{Answer}}}$

$\large{\text{\underline{Note:-}}}$

Firstly,

$\tan\theta=\dfrac{\sin x}{\cos x}$

And,

$\csc x=\dfrac{1}{\sin x}$ and $\sec x=\dfrac{1}{\cos x}$

$\large{\text{\underline{Given to prove:-}}}$

$\hookrightarrow \tan^{2} \beta \times \csc^{2} \beta =\sec^{2} \beta$

According to above,

$\hookrightarrow \text{(L.H.S)}=\dfrac{\sin^{2} x}{\cos^{2} x}\times \dfrac{1}{\sin^{2} x}$

$\hookrightarrow \text{(L.H.S)}=\dfrac{1}{\cos^{2} x}$

$\hookrightarrow \text{(L.H.S)}=\sec^{2} \beta$

Hence,

$\hookrightarrow \text{(L.H.S)}=\text{(R.H.S)}$

Hence proven.

Answer 2

To prove:

$\sf{ {tan}^{2} \beta \times \: {csc}^{2} \beta = {sec}^{2} \beta }$

Solution:

$\sf{LHS =\frac{sin^2x}{cos^2x} \times\frac{1}{sin^2x}}$

$\sf{=\frac{1}{cos^2x}}$

$\sf{ = {sec}^{2} \beta }$

$\sf{ = RHS}$

Hence, proved!!