Question

cosec^2 x-cot^2 y =1 then the value of cos(x-y)​

Answer 1

The value of cos (x-y) is zero when cosec²x - cot²x = 1

Step-by-step explanation:

Given Data

cosec²x - cot² y = 1

To find cos (x-y)

$\begin{aligned}&\csc ^{2} x-\cot ^{2} y=1\\&\frac{1}{\sin ^{2} x}-\frac{\cos ^{2} y}{\sin ^{2} y}=1\end{aligned}$

$\begin{aligned}&\text { Substitute } \sin ^{2} x=1-\cos ^{2} x \text { and } \sin ^{2} y=1-\cos ^{2} y \text { in above equation }\\&\frac{1}{1-\cos ^{2} x}-\frac{\cos ^{2} y}{1-\cos ^{2} y}=1\end{aligned}$

$\begin{aligned}&\frac{1-\cos ^{2} y-\cos ^{2} y\left(1-\cos ^{2} x\right)}{\left(1-\cos ^{2} x\right)\left(1-\cos ^{2} y\right)}=1\\&1-\cos ^{2} y-\cos ^{2} y+\cos ^{2} x \cos ^{2} y=\left(1-\cos ^{2} x\right)\left(1-\cos ^{2} y\right)\\&1-2 \cos ^{2} y+\cos ^{2} x \cos ^{2} y=1-\cos ^{2} x+\cos ^{2} x \cos ^{2} y-\cos ^{2} y\end{aligned}$

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

$-2 \cos ^{2} y+\cos ^{2} y=-\cos ^{2} x\\-\cos ^{2} y+\cos ^{2} x=0$

$\cos ^{2} x-\cos ^{2} y=0\\ \cos ^{2}(x-y)=0$

To learn more ...

1. https://brainly.in/question/4805585

2. https://brainly.in/question/4830919