Question

If the value of trigonometric expression $\cos \left( {{{\tan }^{ - 1}}x + {{\cot }^{ - 1}}\sqrt 3 } \right) = 0$, find the value of x.

Answer 1

Hint – In this question first convert the inner ${\cot ^{ - 1}}\sqrt 3$ in terms of ${\tan ^{ - 1}}\sqrt 3$ using the identity that ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$. The use identity that $\cos \left( {\dfrac{\pi }{2} + \theta } \right) = - \sin \theta$. Since ${\sin ^{ - 1}}0$ is zero hence use this to get the value of x.

Complete step-by-step answer:
As we know that
${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$
$\Rightarrow {\cot ^{ - 1}}x = \dfrac{\pi }{2} - {\tan ^{ - 1}}x$
So use this property in given equation we have,
$\Rightarrow \cos \left( {{{\tan }^{ - 1}}x + \dfrac{\pi }{2} - {{\tan }^{ - 1}}\sqrt 3 } \right) = 0$
$\Rightarrow \cos \left( {\dfrac{\pi }{2} + {{\tan }^{ - 1}}x - {{\tan }^{ - 1}}\sqrt 3 } \right) = 0$
Now as we know that $\cos \left( {\dfrac{\pi }{2} + \theta } \right) = - \sin \theta$ so use this property in above equation we have,
$\Rightarrow \cos \left( {\dfrac{\pi }{2} + {{\tan }^{ - 1}}x - {{\tan }^{ - 1}}\sqrt 3 } \right) = - \sin \left( {{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}\sqrt 3 } \right) = 0$
$\Rightarrow \sin \left( {{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}\sqrt 3 } \right) = 0$
$\Rightarrow {\tan ^{ - 1}}x - {\tan ^{ - 1}}\sqrt 3 = {\sin ^{ - 1}}0$
Now as we know that the value of ${\sin ^{ - 1}}0$ is zero.
So substitute this value in above equation we have,
$\Rightarrow {\tan ^{ - 1}}x - {\tan ^{ - 1}}\sqrt 3 = 0$
$\Rightarrow {\tan ^{ - 1}}x = {\tan ^{ - 1}}\sqrt 3$
Now on comparing we have,
$\Rightarrow x = \sqrt 3$
So this is the required value of x.

Note – It is always advisable to remember the direct trigonometry and trigonometric inverse identities some of them are being mentioned above as they help saving a lot of time. Some other important identities involve ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$, ${\sin ^2}x + {\cos ^2}x = 1$, $1 + {\tan ^2}x = {\sec ^2}x$.