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Question: Find the value of x for the given equation: \(x\cot \left( {90 + \theta } \right) + \tan \left( {9...

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

Explanation

Solution

Hint – In this question use the basic trigonometric conversions likecot⁡(90+θ)=−tan⁡θ, tan⁡(90+θ)=−cot⁡θ and  cos⁡ec(90+θ)=sec⁡θ\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 liketanθ=sin⁡θcos⁡θ{\text{tan}}\theta = \dfrac{{\sin \theta }}{{\cos \theta }}. Use these to find x.

Complete step-by-step answer:
Given trigonometric equation
xcot⁡(90+θ)+tan⁡(90+θ)sin⁡θ+cos⁡ec(90+θ)=0x\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⁡(90+θ)=−tan⁡θ, tan⁡(90+θ)=−cot⁡θ and  cos⁡ec(90+θ)=sec⁡θ\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,
⇒x(−tan⁡θ)+(−cot⁡θ)sin⁡θ+sec⁡(θ)=0\Rightarrow x\left( { - \tan \theta } \right) + \left( { - \cot \theta } \right)\sin \theta + \sec \left( \theta \right) = 0

Now as we know (tan⁡θ=sin⁡θcos⁡θ,cot⁡θ=cos⁡θsin⁡θ,sec⁡θ=1cos⁡θ)\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,
⇒x(−sin⁡θcos⁡θ)+(−cos⁡θsin⁡θ)sin⁡θ+1cos⁡θ=0\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,
⇒−xsin⁡θcos⁡θ−cos⁡θ+1cos⁡θ=0\Rightarrow - x\dfrac{{\sin \theta }}{{\cos \theta }} - \cos \theta + \dfrac{1}{{\cos \theta }} = 0
⇒xsin⁡θcos⁡θ=1cos⁡θ−cos⁡θ\Rightarrow x\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{1}{{\cos \theta }} - \cos \theta
⇒xsin⁡θcos⁡θ=1−cos⁡2θcos⁡θ\Rightarrow x\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{1 - {{\cos }^2}\theta }}{{\cos \theta }}

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

Now cancel out the common terms we have,
⇒x=sin⁡θ\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.