Question

2. Show.
Sec2x + cosec2x=sec2x*cosec2x​

Answer 1

$\underline{\textsf{To show:}}$

$\mathsf{sec^2x+cosec^2x=sec^2x\;cosec^2x}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{sec^2x+cosec^2x}$

$\mathsf{=\dfrac{1}{cos^2x}+\dfrac{1}{sin^2x}}$

$\mathsf{=\dfrac{sin^2x+cos^2x}{cos^2x\;sin^2x}}$

$\mathsf{=\dfrac{1}{cos^2x\;sin^2x}}$

$\mathsf{=\dfrac{1}{cos^2x}\times\dfrac{1}{sin^2x}}$

$\mathsf{=sec^2x{\times}cosec^2x}$

$\implies\boxed{\bf\mathsf{sec^2x+cosec^2x=sec^2x\;cosec^2x}}$

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Answer 2

Given:

Sec2x + cosec2x = sec2x * cosec2x​

To find:

To prove Sec2x + cosec2x = sec2x * cosec2x​

Solution:

From given, we have,

Sec²x + cosec²x = sec²x × cosec²x

​Now consider the LHS part of the given equation.

Sec²x + cosec²x

= 1/cos²x + 1/sin²x

= (sin²x + cos²) / (cosx × sinx)

= 1 / (cosx × sinx)

= 1/cosx × 1/sinx

= sec x × cosec x

Therefore, its proved that Sec²x + cosec²x = sec²x × cosec²x​