Question

Pls answer the question in the attachment​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the Concept of Trigonometry has been used. We see that we are given the question. The correct question is given below . Now we need to prove that LHS = RHS. Firstly we can simplify the LHS and then RHS. Then we can equate both and find the answer.

Let's do it !!

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★ Correct Question :-

Prove that,

$\\\;\bf{\mapsto\;\;\green{\dfrac{(1\;-\;\tan\theta)}{(1\;+\;\tan\theta)}\;=\;\dfrac{(\cot\theta\;+\;1)}{(\cot\theta\;-\;1)}}}$

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★ Solution :-

We know that,

$\\\;\odot\;\;\sf{\tan\theta\;=\;\red{\dfrac{\sin\theta}{\cos\theta}}}$

$\\\;\odot\;\;\sf{\cot\theta\;=\;\red{\dfrac{\cos\theta}{\sin\theta}}}$

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~ For simplification of LHS ::

$\\\;\sf{:\rightarrow\;\;LHS\;=\;\bf{\dfrac{(1\;-\;\tan\theta)}{(1\;+\;\tan\theta)}}}$

By applying the values,

$\\\;\sf{:\Longrightarrow\;\;\dfrac{\bigg(1\;-\;\dfrac{\sin\theta}{\cos\theta}\bigg)}{\bigg(1\;+\;\dfrac{\sin\theta}{\cos\theta}\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;\dfrac{\bigg(\dfrac{\cos\theta\;-\;\sin\theta}{\cos\theta}\bigg)}{\bigg(\dfrac{\cos\theta\;+\;\sin\theta}{\cos\theta}\bigg)}}$

Cancelling the like term, we get

$\\\;\sf{:\Longrightarrow\;\;\bigg(\dfrac{\cos\theta\;-\;\sin\theta}{\cos\theta\;+\;\sin\theta}\bigg)}$

$\\\;\bf{:\Longrightarrow\;\;\orange{\dfrac{\cos\theta\;-\;\sin\theta}{\cos\theta\;+\;\sin\theta}}}$

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~ For simplification of RHS ::

$\\\;\sf{:\rightarrow\;\;RHS\;=\;\dfrac{(\cot\theta\;+\;1)}{(\cot\theta\;-\;1)}}$

By applying values, we get

$\\\;\sf{:\Longrightarrow\;\;\dfrac{\bigg(\dfrac{\cos\theta}{\sin\theta}\;-\;1\bigg)}{\bigg(\dfrac{\cos\theta}{\sin\theta}\;+\;1\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;\dfrac{\bigg(\dfrac{\cos\theta\;-\;\sin\theta}{\sin\theta}\bigg)}{\bigg(\dfrac{\cos\theta\;-\;\sin\theta}{\sin\theta}\bigg)}}$

Cancelling the like terms, we get

$\\\;\sf{:\Longrightarrow\;\;\bigg(\dfrac{\cos\theta\;-\;\sin\theta}{\cos\theta\;+\;\sin\theta}\bigg)}$

$\\\;\bf{:\Longrightarrow\;\;\orange{\dfrac{\cos\theta\;-\;\sin\theta}{\cos\theta\;+\;\sin\theta}}}$

Now equating both LHS = RHS, we get

$\\\;\bf{:\mapsto\;\;\blue{\dfrac{\cos\theta\;-\;\sin\theta}{\cos\theta\;+\;\sin\theta}\;=\;\dfrac{\cos\theta\;-\;\sin\theta}{\cos\theta\;+\;\sin\theta}}}$

Clearly LHS = RHS.

$\\\;\qquad\quad\underline{\boxed{\bf{\purple{Hence,\;\;Proved}}}}$

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★ More Formulas to know :-

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

$\\\;\tt{\leadsto\;\;cosec^{2}\theta\;=\;1\;-\;\tan^{2}\theta}$

$\\\;\tt{\leadsto\;\;\sec^{2}\theta\;=\;1\;-\;\cot^{2}\theta}$

$\\\;\tt{\leadsto\;\;cosec\theta\;=\;\dfrac{1}{\sin\theta}}$

$\\\;\tt{\leadsto\;\;\sec\theta\;=\;\dfrac{1}{\cos\theta}}$

$\\\;\tt{\leadsto\;\;\tan\theta\;=\;\dfrac{1}{\cot\theta}}$