Question

plz solve this｡:ﾟ(;´∩`;)ﾟ:｡

don't spam :(​

Answer 1

proof of above question ....

Answer 2

★ Concept :-

Here the concept of Trigonometric Identities has been used. We see that we are given a relationship. From that we can derive amother relationship. Then we can apply the formula of tan(x - y) and thus by applying different Trigonometric Identities we can get answer.

Let's do it !!

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★ Formula Used :-

$\;\boxed{\sf{\pink{\tan(x\:-\:y)\;=\;\dfrac{\tan x\:-\:\tan y}{1\:+\:\tan x\tan y}}}}$

$\;\boxed{\sf{\pink{\sin2\theta\;=\;2\sin\theta\cos\theta}}}$

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★ Solution :-

Given,

$\;\bf{\mapsto\;\;\green{2\tan\alpha\;=\;3\tan\beta}}$

From this we can get,

$\;\bf{\mapsto\;\;\green{\tan\alpha\;=\;\dfrac{3}{2}\tan\beta}}$

Let this be equation i) .

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~ For proving the given equation ::

We know that,

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

• Here x = α

• Here y = β

By applying these values, we get

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

Now we need to convert this equation in same terms that is in terms of either α or β. Let's use the term β here for convenience.

By applying the value of equation i) we get,

$\;\sf{\rightarrow\;\;\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{\dfrac{3}{2}\tan\beta\:-\:\tan\beta}{1\:+\:\dfrac{3}{2}\tan\beta\tan\beta}}}$

On solving this, we get

$\;\sf{\rightarrow\;\;\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{\dfrac{3\tan\beta\:-\:2\tan\beta}{2}}{1\:+\:\dfrac{3}{2}\tan^{2}\beta}}}$

$\;\sf{\rightarrow\;\;\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{\dfrac{\tan\beta}{2}}{1\:+\:\dfrac{3}{2}\tan^{2}\beta}}}$

Multiplying 2 to both the numerator and denominator, we get

$\;\sf{\rightarrow\;\;\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{2\:\times\:\dfrac{\tan\beta}{2}}{2\:\times\:\bigg(1\:+\:\dfrac{3}{2}\tan^{2}\beta\bigg)}}}$

$\;\sf{\rightarrow\;\;\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{\tan\beta}{2\:+\:3\tan^{2}\beta}}}$

Now dividing multiplying both numerator and denominator by cos²β we get,

$\;\sf{\rightarrow\;\;\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{\tan\beta\:\times\:\cos^{2}\beta}{\cos^{2}\beta(2\:+\:3\tan^{2}\beta)}}}$

We know that tanθ = sinθ/cosθ. By applying this, we get

$\;\sf{\rightarrow\;\;\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{\dfrac{\sin\beta}{\cos\beta}\:\times\:\cos^{2}\beta}{2\cos^{2}\beta\:+\:3\bigg(\cos^{2}\beta\:\times\:\dfrac{\sin^{2}\beta}{\cos^{2}\beta}\bigg)}}}$

Cancelling the like terms we get,

$\;\sf{\rightarrow\;\;\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{\sin\beta\cos\beta}{2\cos^{2}\beta\:+\:3\sin^{2}\beta}}}$

Now multiplying both numerator and denominator by 2 we get,

$\;\sf{\rightarrow\;\;\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{2(\sin\beta\cos\beta)}{2(2\cos^{2}\beta\:+\:3\sin^{2}\beta)}}}$

$\;\sf{\Longrightarrow\;\;\orange{\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{2\sin\beta\cos\beta}{4\cos^{2}\beta\:+\:6\sin^{2}\beta}}}}$

We know that,

$\;\tt{\blue{\odot\;\;\sin2\theta\;=\;2\sin\theta\cos\theta}}$

By applying this, we get

$\;\sf{\Longrightarrow\;\;\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{\sin 2\beta}{4\cos^{2}\beta\:+\:6\sin^{2}\beta}}}$

This will give us,

$\;\sf{\Longrightarrow\;\;\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{\sin 2\beta}{4\:+\:(1\:-\:\cos 2\beta)}}}$

$\;\sf{\Longrightarrow\;\;\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{\sin 2\beta}{4\:+\:1\:-\:\cos 2\beta}}}$

$\;\sf{\Longrightarrow\;\;\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{\sin 2\beta}{5\:-\:\cos 2\beta}}}$

This is the thing to be proved.

$\;\bf{\Longrightarrow\;\;\purple{\tan(\alpha\:-\:\beta)\;=\;\bf{\dfrac{\sin 2\beta}{5\:-\:\cos 2\beta}}}}$

Hence proved.

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★ More to know :-

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

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

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

$\;\tt{\leadsto\;\;\sin(-x)\;=\;-\sin x}$

$\;\tt{\leadsto\;\;\cos(-x)\;=\;\cos x}$