# Prove the following trigonometric equation

$\dfrac{{\sin \left( {{{90}^0} - \theta } \right)}}{{\cos ec\left( {{{90}^0} - \theta } \right) - \cot \left( {{{90}^0} - \theta } \right)}} = 1 + \sin \theta $

Last updated date: 26th Mar 2023

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Answer

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

Given that

$\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

$ = \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 \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

$ \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 \theta $ so, we will substitute the value of $\sec \theta \& \tan \theta $ in terms of $\sin \theta \& \cos \theta $ .

$\because \sec \theta = \dfrac{1}{{\cos \theta }}\& \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$

After putting the value of $\sec \theta \& \tan \theta $ in the given equation, we obtain

$

\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 ${\sin ^2}\theta + {\cos ^2}\theta = 1$

$

\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

$

\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.

Given that

$\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

$ = \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 \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

$ \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 \theta $ so, we will substitute the value of $\sec \theta \& \tan \theta $ in terms of $\sin \theta \& \cos \theta $ .

$\because \sec \theta = \dfrac{1}{{\cos \theta }}\& \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$

After putting the value of $\sec \theta \& \tan \theta $ in the given equation, we obtain

$

\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 ${\sin ^2}\theta + {\cos ^2}\theta = 1$

$

\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

$

\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.

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