If $cosec^2 x-cot^2 y=1$, then what is the value of $cos^2(x-y)$?

Given :

$cosec^2 x-cot^2 y=1$.

To do :

We have to find the value of $cos^2 (x-y)$.

Solution :

We know that,

$cosec x = \frac{1}{sin x}$ and $cot y = \frac{cos y}{sin y}$

$cosec^2 x-cot^2 y=1$

 $\frac{1}{sin^2 x} - \frac{cos^2 y}{sin^2 y} = 1$

Substitute $sin^2 x = 1-cos^2 x$ and $sin^2 y = 1-cos^2 y$ in the above equation,

$\frac{1}{ 1-cos^2 x} - \frac{cos^2y}{ 1-cos^2 y} = 1$

$ \frac{1-cos^2 y-cos^2 y(1-cos^2 x)}{(1-cos^2 x) (1-cos^2 y)} = 1$

$1-cos^2 y-cos^2 y+cos^2 x cos^2 y = (1-cos^2 x) (1-cos^2 y)$

$1-2cos^2 y+cos^2 x cos^2 y = 1- cos^2 x + cos^ x cos^2 y - cos^2 y$

$1-2cos^2 y = 1- cos^2 x -cos^2 y$

Also '1' can be cancelled on both sides $-2cos^2 y = -cos^2 x -cos^2 y$

$-2cos^2 y +cos^2 y=- cos^2 x $

$- cos^2 y+ cos^2 x = 0$

$ cos^2 x - cos^2 y=0$

$cos^2 (x-y)=0$

Therefore, the value of $cos^2 (x-y)$ is 0.

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Updated on: 10-Oct-2022

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