Solveeit Logo
Login

Question

Question: Prove the following trigonometric equation \(\dfrac{{\sin \left( {{{90}^0} - \theta } \right)}}{{\...

Prove the following trigonometric equation
sin⁡(900−θ)cos⁡ec(900−θ)−cot⁡(900−θ)=1+sin⁡θ\dfrac{{\sin \left( {{{90}^0} - \theta } \right)}}{{\cos ec\left( {{{90}^0} - \theta } \right) - \cot \left( {{{90}^0} - \theta } \right)}} = 1 + \sin \theta

Explanation

Solution

Hint: For solving such equations use general trigonometric identities of angle transformation such as sin⁡(900−θ)=cos⁡θ\sin \left( {{{90}^0} - \theta } \right) = \cos \theta and proceed further by simplifying the equation.

Given that
sin⁡(900−θ)cos⁡ec(900−θ)−cot⁡(900−θ)=1+sin⁡θ\dfrac{{\sin \left( {{{90}^0} - \theta } \right)}}{{\cos ec\left( {{{90}^0} - \theta } \right) - \cot \left( {{{90}^0} - \theta } \right)}} = 1 + \sin \theta
We proceed further by taking the LHS side
=sin⁡(900−θ)cos⁡ec(900−θ)−cot⁡(900−θ)= \dfrac{{\sin \left( {{{90}^0} - \theta } \right)}}{{\cos ec\left( {{{90}^0} - \theta } \right) - \cot \left( {{{90}^0} - \theta } \right)}}
As we know that
sin⁡(900−θ)=cos⁡θ cos⁡ec(900−θ)=sec⁡θ cot⁡(900−θ)=tan⁡θ  \sin \left( {{{90}^0} - \theta } \right) = \cos \theta \\\ \cos ec\left( {{{90}^0} - \theta } \right) = \sec \theta \\\ \cot \left( {{{90}^0} - \theta } \right) = \tan \theta \\\
So after substituting these terms in given equation we get
⇒sin⁡(900−θ)cos⁡ec(900−θ)−cot⁡(900−θ)=cos⁡θsec⁡θ−tan⁡θ\Rightarrow \dfrac{{\sin \left( {{{90}^0} - \theta } \right)}}{{\cos ec\left( {{{90}^0} - \theta } \right) - \cot \left( {{{90}^0} - \theta } \right)}} = \dfrac{{\cos \theta }}{{\sec \theta - \tan \theta }}
Since we need the RHS in terms of sin⁡θ\sin \theta so, we will substitute the value of sec⁡θ&tan⁡θ\sec \theta \& \tan \theta in terms of sin⁡θ&cos⁡θ\sin \theta \& \cos \theta .
∵sec⁡θ=1cos⁡θ&tan⁡θ=sin⁡θcos⁡θ\because \sec \theta = \dfrac{1}{{\cos \theta }}\& \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
After putting the value of sec⁡θ&tan⁡θ\sec \theta \& \tan \theta in the given equation, we obtain
⇒cos⁡θsec⁡θ−tan⁡θ=cos⁡θ1cos⁡θ−sin⁡θcos⁡θ =cos⁡θ1−sin⁡θcos⁡θ =cos⁡2θ1−sin⁡θ  \Rightarrow \dfrac{{\cos \theta }}{{\sec \theta - \tan \theta }} = \dfrac{{\cos \theta }}{{\dfrac{1}{{\cos \theta }} - \dfrac{{\sin \theta }}{{\cos \theta }}}} \\\ = \dfrac{{\cos \theta }}{{\dfrac{{1 - \sin \theta }}{{\cos \theta }}}} \\\ = \dfrac{{{{\cos }^2}\theta }}{{1 - \sin \theta }} \\\
As we know that sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
⇒cos⁡2θ=1−sin⁡2θ =(1+sin⁡θ)(1−sin⁡θ)[∵a2−b2=(a+b)(a−b)]  \Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta \\\ = \left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)\left[ {\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right] \\\

Substituting in the given term, we have
⇒cos⁡2θ1−sin⁡θ=(1+sin⁡θ)(1−sin⁡θ)1−sin⁡θ =(1+sin⁡θ)  \Rightarrow \dfrac{{{{\cos }^2}\theta }}{{1 - \sin \theta }} = \dfrac{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}}{{1 - \sin \theta }} \\\ = \left( {1 + \sin \theta } \right) \\\
This is the same as RHS.

Hence the equation is proved.

Note: In order to solve such questions involving different trigonometric terms always first try to simplify the angle of the terms and then use trigonometric terms to further simplify the terms. In order to fetch the result on the other side or to prove some terms, always keep in mind the terms on the other side while making any substitution as some substitution may further make the term complex.