Question

please answer this question ​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of Trigonometric Identities has been used. We are given a equation and we have to prove that LHS = RHS. Then firstly we will simplify the LHS of the equation. This can be done by reducing the terms. For this we will multiply the terms and then substitute the values to find the answer.

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★ Identities Used :-

$\\\;\boxed{\sf{\sin^{2}\:\theta\;=\;\bf{1\;-\;\cos^{2}\:\theta}}}$

$\\\;\boxed{\sf{\cosec^{2}\:\theta\;=\;\bf{1\;+\;\cot^{2}\:\theta}}}$

$\\\;\boxed{\sf{\cosec\:\theta\;=\;\bf{\dfrac{1}{\sin\:\theta}}}}$

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★ To Prove :-

$\\\sf{\odot\;\;\quad\;(1\;+\;\cos\:\theta)\:(1\;-\;\cos\:\theta)\:(1\;+\;\cot^{2}\:\theta)\;\;=\;\;1}$

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★ Solution :-

Given,

• LHS ::

✒ (1 + cosθ)(1 - cosθ)(1 + cot²θ)

• RHS ::

✒ 1

Now firstly let's simply LHS.

Then,

~ For the value of LHS ::

$\\\;\sf{:\rightarrow\;\;L.H.S.\;=\;\bf{(1\;+\;\cos\:\theta)\:(1\;-\;\cos\:\theta)\:(1\;+\;\cot^{2}\:\theta)}}$

Multiplying first two terms, we get,

$\\\;\sf{:\Longrightarrow\;L.H.S.\;=\;\sf{\bigg(1\;-\;\cos\:\theta\;+\;\cos\:\theta\;-\;\cos^{2}\:\theta\bigg)\:(1\;+\;\cot^{2}\:\theta)}}$

$\\\;\sf{:\Longrightarrow\;L.H.S.\;=\;\sf{(1\;-\;\cos^{2}\:\theta)\:(1\;+\;\cot^{2}\:\theta)}}$

Using the first identity here that ::

→ sin² θ = 1 - cos² θ we get,

$\\\;\sf{:\Longrightarrow\;L.H.S.\;=\;\sf{(\sin^{2}\:\theta)\:(1\;+\;\cot^{2}\:\theta)}}$

$\\\;\sf{:\Longrightarrow\;L.H.S.\;=\;\sf{\sin^{2}\:\theta\;\times\;(1\;+\;\cot^{2}\:\theta)}}$

Now using the second identity here that ::

→ cosec² θ = 1 + cot² θ we get,

$\\\;\sf{:\Longrightarrow\;L.H.S.\;=\;\sf{\sin^{2}\:\theta\;\times\;(cosec^{2}\:\theta)}}$

$\\\;\sf{:\Longrightarrow\;L.H.S.\;=\;\sf{\sin^{2}\:\theta\;\times\;cosec^{2}\:\theta}}$

Also, we know that,

$\\\;\sf{:\rightarrow\;\;cosec\:\theta\;=\;\bf{\dfrac{1}{\sin\:\theta}}}$

Now using this in the equation of LHS, we get,

$\\\;\sf{:\Longrightarrow\;L.H.S.\;=\;\bf{\sin^{2}\:\theta\;\times\;\bigg(\dfrac{1}{\sin\:\theta}\bigg)^{2}}}$

$\\\;\sf{:\Longrightarrow\;L.H.S.\;=\;\bf{\sin^{2}\:\theta\;\times\;\bigg(\dfrac{1}{\sin^{2}\:\theta}\bigg)}}$

Cancelling sin² θ, we get,

$\\\;\sf{:\Longrightarrow\;L.H.S.\;=\;\bf{\cancel{\sin^{2}\:\theta}\;\times\;\bigg(\dfrac{1}{\cancel{\sin^{2}\:\theta}}\bigg)}}$

$\\\;\sf{:\Longrightarrow\;L.H.S.\;=\;\bf{1\;\times\;1}}$

$\\\;\bf{:\Longrightarrow\;L.H.S.\;=\;\bf{1}}$

And, we know that RHS = 1. So,

$\\\;\bf{\red{:\mapsto\;\;L.H.S.\;=\;RHS\;=\;\bf{1}}}$

$\\\;\qquad\quad\boxed{\underline{\tt{\blue{Hence,\;\;Proved}}}}$

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_____________________________________________★ More Formulas to know :-

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

$\\\;\sf{\leadsto\;\;\sec^{2}\theta\;=\;1\;+\;\tan^{2}\theta}$

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

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

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