Question

Show that

Answer 1

\[\cos(\tan ^{-1}[\sin(\cot ^{-1}x)]) = \sqrt{\frac{ x^2+1 }{ x^2+2 }} \]

Answer 2

i need paper and pen for that :P

Answer 3

\[\cos(\tan ^{-1}[\sin(\cot ^{-1}x)]) = \sqrt{\frac{ x^2-1 }{ x^2-2 }}\]

Answer 4

was there a typo in +- signs ?

Answer 5

both should be -

Answer 6

\[=\sqrt{\frac{ x^2+1 }{ x^2+2 }}\]

Answer 7

oh yes yes...

Answer 8

il do it the long way maybe :

say \(\cot \alpha = x \implies \sin \alpha = \sqrt{\dfrac{1}{1+x^2}}\)

Answer 9

\(\cos(\tan ^{-1}[\sin(\cot ^{-1}x)]) = \cos (\tan^{-1} \left(\sqrt{\dfrac{1}{1+x^2}}\right))\)

Answer 10

Next, say \(\tan \beta = \sqrt{\dfrac{1}{1+x^2}} \implies \cos \beta = \sqrt{\dfrac{1+x^2}{2+x^2}}\)

Answer 11

not so much of an exciting problem @iambatman ... just involves drawing the triangle two times :)

Answer 12

Yeah I see where it's leading to lol, don't worry about it :d

Answer 13

Night... !

Answer 14

Good night :3

Answer 15

night ?
i thought you both are from india :P

Answer 16

@hartnn I'm from Canada :). And it's very late here >.<