Question

root sec sq theta.+root cosec sq. theta=tan theta+cot theta prove it​

Answer 1

Explanation:

its your answer is ready

Answer 2

$\blue\bigstar$QUESTION:-

Prove that

$\to \bf \sqrt{sec^{2} \theta + cosec ^{2} \theta} = tan \theta + cot \theta$

$\red\bigstar$PROOF:-

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We have

$\bf LHS= \sqrt{ {sec}^{2} \theta+ {cosec}^{2} \theta}$

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Using below trignometric identity

$\boxed{ \bf \frac{1}{ {cos}\theta } = sec \theta, \frac{1}{sin \theta}=cosec \theta}$

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We have ,

$\implies \bf \sqrt{ \dfrac{1}{ {cos}^{2} \theta } + \dfrac{1}{ {sin}^{2} \theta } }$

$\implies \bf \sqrt{ \dfrac{ {sin}^{2 } \theta + {cos}^{2} \theta }{sin^{2} \theta \: cos^{2} \theta} }$

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Using below trignometric identity

$\boxed { \bf {sin}^{2} \theta + cos^{2}\theta = 1}$

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We have

$\implies \bf\sqrt{ \dfrac{1}{ {sin}^{2} \theta \: cos^{2} \theta} }$

$\implies \bf \dfrac{ \sqrt{1} }{ \sqrt{sin^{2} \theta \: cos^{2} \theta} }$

$\implies \bf \dfrac{1}{sin \theta \: cos \theta}$

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Again using below trignometric identity

$\boxed{ \bf sin^{2} \theta + cos ^{2} \theta = 1 }$

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We have

$\implies \bf \dfrac{ {sin}^{2} \theta + cos^{2} \theta}{sin \theta \times cos \theta}$

$\implies \bf \dfrac{ {sin}^{2} \theta}{sin \theta \: cos \theta} + \dfrac{ {cos}^{2} \theta}{sin \theta \: cos \theta}$

$\implies \bf\dfrac{ \cancel{sin\theta }\times sin \theta}{ \cancel{sin \theta } \times \: cos \theta} +\dfrac{ cos \theta \times \cancel{ cos \theta}}{sin \theta \times \cancel{cos \theta}}$

$\implies \bf \dfrac{sin \theta}{cos \theta} + \dfrac{cos \theta}{sin \theta}$

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Using below trignometric identity

$\boxed {\bf \frac{sin \theta}{cos \theta} = tan \theta \: , \frac{cos \theta}{sin \theta } = cot \theta}$

We have

$\implies \bf tan \theta + cos \theta$

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=RHS

Therefore,LHS=RHS .......Proved