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Question: Find the value of \(\dfrac{{{{\cot }^2}15^\circ - 1}}{{{{\cot }^2}15^\circ + 1}} = \) A. \(\dfrac{...

Find the value of cot⁡215∘−1cot⁡215∘+1=\dfrac{{{{\cot }^2}15^\circ - 1}}{{{{\cot }^2}15^\circ + 1}} =
A. 12\dfrac{1}{2}
B. 32\dfrac{{\sqrt 3 }}{2}
C. 3343\dfrac{{\sqrt 3 }}{4}
D. 3\sqrt 3

Explanation

Solution

First, we need to analyze the given information so that we can able to solve the problem. Generally, in Mathematics, the trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
Here in this question, we are asked to find the value of cot⁡215∘−1cot⁡215∘+1\dfrac{{{{\cot }^2}15^\circ - 1}}{{{{\cot }^2}15^\circ + 1}}
We need to apply the appropriate trigonometric identities to obtain the required answer.
Formula to be used:
cot⁡θ=1tan⁡θ\cot \theta = \dfrac{1}{{\tan \theta }}
cos⁡2θ=1−tan⁡2θ1+tan⁡2θ\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}

Complete step by step answer:
We are asked to calculate the value of cot⁡215∘−1cot⁡215∘+1\dfrac{{{{\cot }^2}15^\circ - 1}}{{{{\cot }^2}15^\circ + 1}}
cot⁡215∘−1cot⁡215∘+1=1tan⁡215∘−11tan⁡215∘+1\dfrac{{{{\cot }^2}15^\circ - 1}}{{{{\cot }^2}15^\circ + 1}} = \dfrac{{\dfrac{1}{{{{\tan }^2}15^\circ }} - 1}}{{\dfrac{1}{{{{\tan }^2}15^\circ }} + 1}}(Here we applied cot⁡θ=1tan⁡θ\cot \theta = \dfrac{1}{{\tan \theta }} )
=1−tan⁡215∘tan⁡215∘1+tan⁡215∘tan⁡215∘= \dfrac{{\dfrac{{1 - {{\tan }^2}15^\circ }}{{{{\tan }^2}15^\circ }}}}{{\dfrac{{1 + {{\tan }^2}15^\circ }}{{{{\tan }^2}15^\circ }}}}
=1−tan⁡215∘tan⁡215∘×tan⁡215∘1+tan⁡215∘= \dfrac{{1 - {{\tan }^2}15^\circ }}{{{{\tan }^2}15^\circ }} \times \dfrac{{{{\tan }^2}15^\circ }}{{1 + {{\tan }^2}15^\circ }}
=1−tan⁡215∘1+tan⁡215∘= \dfrac{{1 - {{\tan }^2}15^\circ }}{{1 + {{\tan }^2}15^\circ }}
=cos⁡2(15∘)= \cos 2\left( {15^\circ } \right) (Here we applied the trigonometric identity cos⁡2θ=1−tan⁡2θ1+tan⁡2θ\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} )
=cos⁡30∘= \cos 30^\circ
=32= \dfrac{{\sqrt 3 }}{2}
Hence, cot⁡215∘−1cot⁡215∘+1=32\dfrac{{{{\cot }^2}15^\circ - 1}}{{{{\cot }^2}15^\circ + 1}} = \dfrac{{\sqrt 3 }}{2}

So, the correct answer is “Option B”.

Note: If we are asked to calculate the value of a trigonometric expression, we need to first analyze the given problem where we are able to apply the trigonometric identities.
Here, we have applied some trigonometric identities/formulae that are needed to know to obtain the desired answer. Hence, we gotcot⁡215∘−1cot⁡215∘+1=32\dfrac{{{{\cot }^2}15^\circ - 1}}{{{{\cot }^2}15^\circ + 1}} = \dfrac{{\sqrt 3 }}{2}.