Question

Find the value of sin (tan-1 99 +cot-1 99) ​

Answer 1

Answer:

we know that

$tan {}^{ - 1} x \: + \: {cot}^{ - 1} x \: = \frac{\pi}{2}$

then

$\sin(tan {}^{ - 1} 99 \: + \: {cot}^{ - 1} 99 \: ) = \sin( \frac{\pi}{2} ) = 1$

Answer 2

Answer:

The value of $sin (tan^{-1} 99 +cot^{-1} 99)=1$

Step-by-step explanation:

Given the trigonometric relation, $sin (tan^{-1} 99 +cot^{-1} 99)$

Let us consider,

$tan^{-1} x =y$                                               ....(1)

$\implies x =tan y$

Using the trigonometric relation, tanθ = cot(90 - θ)

$\implies x =cot(90-y)$

$\implies cot^{-1}x =90-y$

Substituting (1) for y,

$\implies cot^{-1}x =90-tan^{-1} x$

$\implies cot^{-1}x +tan^{-1} x =90$

Using this in the given relation, $tan^{-1} 99 +cot^{-1} 99= 90$

And hence,

$sin (tan^{-1} 99 +cot^{-1} 99)=sin 90$

The value of sin 90 = 1,

Therefore, the value of $sin (tan^{-1} 99 +cot^{-1} 99)=1$