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Question: How do you prove: \[\dfrac{{\cos \theta }}{{1 - \tan \theta }} + \dfrac{{\sin \theta }}{{1 - \cot \t...

How do you prove: cos⁡θ1−tan⁡θ+sin⁡θ1−cot⁡θ=2sin⁡(45∘+θ)\dfrac{{\cos \theta }}{{1 - \tan \theta }} + \dfrac{{\sin \theta }}{{1 - \cot \theta }} = \sqrt 2 \sin \left( {{{45}^ \circ } + \theta } \right) ?

Explanation

Solution

Hint : The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{{\sin x}}{{\cos x}} and sec⁡x=1cos⁡x\sec x = \dfrac{1}{{\cos x}} . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem and proving the result given to us.

Complete step by step solution:
In the given problem, we have to prove a trigonometric identity that can be further used in many questions and problems as a direct result and has wide ranging applications. For proving the desired result, we need to first know the definitions of all the six trigonometric ratios.
Now, we need to make the left and right sides of the equation equal.
L.H.S. =cos⁡θ1−tan⁡θ+sin⁡θ1−cot⁡θ = \dfrac{{\cos \theta }}{{1 - \tan \theta }} + \dfrac{{\sin \theta }}{{1 - \cot \theta }}
As we know that tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{{\sin x}}{{\cos x}} and sec⁡x=1cos⁡x\sec x = \dfrac{1}{{\cos x}} . So, we get,
=cos⁡θ1−(sin⁡θcos⁡θ)+sin⁡θ1−(cos⁡θsin⁡θ)= \dfrac{{\cos \theta }}{{1 - \left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)}} + \dfrac{{\sin \theta }}{{1 - \left( {\dfrac{{\cos \theta }}{{\sin \theta }}} \right)}}
Taking LCM of fractions, we get,
=cos⁡θ(cos⁡θ−sin⁡θcos⁡θ)+sin⁡θ(sin⁡θ−cos⁡θsin⁡θ)= \dfrac{{\cos \theta }}{{\left( {\dfrac{{\cos \theta - \sin \theta }}{{\cos \theta }}} \right)}} + \dfrac{{\sin \theta }}{{\left( {\dfrac{{\sin \theta - \cos \theta }}{{\sin \theta }}} \right)}}
=cos⁡2θ(cos⁡θ−sin⁡θ)+sin⁡2θ(sin⁡θ−cos⁡θ)= \dfrac{{{{\cos }^2}\theta }}{{\left( {\cos \theta - \sin \theta } \right)}} + \dfrac{{{{\sin }^2}\theta }}{{\left( {\sin \theta - \cos \theta } \right)}}
Taking negative sign common from the last term, we get,
=cos⁡2θ(cos⁡θ−sin⁡θ)−sin⁡2θ(cos⁡θ−sin⁡θ)= \dfrac{{{{\cos }^2}\theta }}{{\left( {\cos \theta - \sin \theta } \right)}} - \dfrac{{{{\sin }^2}\theta }}{{\left( {\cos \theta - \sin \theta } \right)}}
=cos⁡2θ−sin⁡2θ(cos⁡θ−sin⁡θ)= \dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{\left( {\cos \theta - \sin \theta } \right)}}
Factorizing the numerator using algebraic identity (a2−b2)=(a−b)(a+b)\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) ,
=(cos⁡θ+sin⁡θ)(cos⁡θ−sin⁡θ)(cos⁡θ−sin⁡θ)= \dfrac{{\left( {\cos \theta + \sin \theta } \right)\left( {\cos \theta - \sin \theta } \right)}}{{\left( {\cos \theta - \sin \theta } \right)}}
=(cos⁡θ+sin⁡θ)= \left( {\cos \theta + \sin \theta } \right)
Multiplying the numerator and denominator by 2\sqrt 2 ,
=2(12cos⁡θ+12sin⁡θ)= \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}\cos \theta + \dfrac{1}{{\sqrt 2 }}\sin \theta } \right)
We know that sin⁡45∘=12\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }} and cos⁡45∘=12\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }} .
=2(sin⁡(45∘)cos⁡θ+cos⁡(45∘)sin⁡θ)= \sqrt 2 \left( {\sin \left( {{{45}^ \circ }} \right)\cos \theta + \cos \left( {{{45}^ \circ }} \right)\sin \theta } \right)
Using sine compound angle formula, we get,
=2sin⁡(45∘+θ)=R.H.S.= \sqrt 2 \sin \left( {{{45}^ \circ } + \theta } \right) = R.H.S.
As the left side of the equation is equal to the right side of the equation, we have,
cos⁡θ1−tan⁡θ+sin⁡θ1−cot⁡θ=2sin⁡(45∘+θ)\dfrac{{\cos \theta }}{{1 - \tan \theta }} + \dfrac{{\sin \theta }}{{1 - \cot \theta }} = \sqrt 2 \sin \left( {{{45}^ \circ } + \theta } \right)

Note : Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations.