Question

Find the value of x for the given equation:
$x\cot \left( {90 + \theta } \right) + \tan \left( {90 + \theta } \right)\sin \theta + cosec\left( {90 + \theta } \right) = 0$

Answer 1

Hint – In this question use the basic trigonometric conversions like$\cot \left( {90 + \theta } \right) = - \tan \theta ,{\text{ }}\tan \left( {90 + \theta } \right) = - \cot \theta {\text{ and}}\;\cos ec\left( {90 + \theta } \right) = \sec \theta $, also some basic trigonometric ratios can be expressed in other ratios like${\text{tan}}\theta = \dfrac{{\sin \theta }}{{\cos \theta }}$. Use these to find x.

Complete step-by-step answer:
Given trigonometric equation
$x\cot \left( {90 + \theta } \right) + \tan \left( {90 + \theta } \right)\sin \theta + \cos ec\left( {90 + \theta } \right) = 0$

Now as we know some of the basic trigonometric properties such as $\cot \left( {90 + \theta } \right) = - \tan \theta ,{\text{ }}\tan \left( {90 + \theta } \right) = - \cot \theta {\text{ and}}\;\cos ec\left( {90 + \theta } \right) = \sec \theta $ so use these properties in above equation we have,
$ \Rightarrow x\left( { - \tan \theta } \right) + \left( { - \cot \theta } \right)\sin \theta + \sec \left( \theta \right) = 0$

Now as we know $\left( {\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }},\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }},\sec \theta = \dfrac{1}{{\cos \theta }}} \right)$ so substitute these values in above equation we have,
$ \Rightarrow x\left( { - \dfrac{{\sin \theta }}{{\cos \theta }}} \right) + \left( { - \dfrac{{\cos \theta }}{{\sin \theta }}} \right)\sin \theta + \dfrac{1}{{\cos \theta }} = 0$

Now simplify the above equation we have,
$ \Rightarrow - x\dfrac{{\sin \theta }}{{\cos \theta }} - \cos \theta + \dfrac{1}{{\cos \theta }} = 0$
$ \Rightarrow x\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{1}{{\cos \theta }} - \cos \theta $
$ \Rightarrow x\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{1 - {{\cos }^2}\theta }}{{\cos \theta }}$

Now as we know $\left( {1 - {{\cos }^2}\theta } \right) = {\sin ^2}\theta $ so substitute this value in above equation we have,
$ \Rightarrow x\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{{{\sin }^2}\theta }}{{\cos \theta }}$

Now cancel out the common terms we have,
$ \Rightarrow x = \sin \theta $

So this is the required value of x for which the given trigonometric equation becomes zero.
So this is the required answer.

Note – It is always advisable to remember such basic identities while involving trigonometric questions as it helps save a lot of time. Eventually it’s difficult to mug up every identity but with practice things get easier, so keep practicing.