Question

if sin x sin y - cos x cos y +1 =0 then prove cot x tan y = -1

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The value of $cot(x) tan( y)=-1$

Step-by-step explanation:

We are given that

$sin(x) sin(y)-cos(x) cos(y)=-1$

$cos(x)cos(y)-sin(x)sin(y)=1$

$cos(x+y)=1$      (by formula of trigonometry)

Take inverse of cos on both sides. we get,

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

Substitute $cos^-1(1)=0$. Then we get,

$x+y=0$               $(cos(0)=1)$

$x=-y$

Substitute -x=y and simplify it.

$cot(x) tan(y)=cot(x) tan(-x)$

tan(-x) can be written as $-tanx$

                    $=cot(x)$×$-tan(x)$

Write $tanx$ as $\frac{1}{cotx}$ to get the required answer.

                     $=-cot(x)$ × $\frac{1}{cot(x)}$

                     $=-1$

Hence proved.