Question

Prove this identity in trigonometry ​

Answer 1

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${(sina + seca})^{2} + ( {cosa + coseca)}^{2} = {(1 + secacoseca)}^{2}$

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Here we take L.H.S to prove this question.

see ⇑⇑⇑in the picture above attach for complete solution step by step:-

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Answer 2

Step-by-step explanation:

Given that:

$( {sinA + secA)}^{2} + ( {cosA + cosecA)}^{2} = ( {1 + secA \: cosecA)}^{2}$

Solution:

To prove this trigonometric expression

It is easy to take LHS

$( {sinA + secA)}^{2} + ( {cosA + cosecA)}^{2}$

Open square using Algebraic identity

$( {a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

$= {sin}^{2} A + {sec}^{2} A + 2sinA \: sec \: A + {cos}^{2} A + {cosec}^{2} A + 2cosA \: cosec \: A$

take terms of sin A and cos A together

$= {sin}^{2} A +{cos}^{2} A + {sec}^{2} A + 2sinA \: sec \: A + {cosec}^{2} A + 2cosA \: cosec \: A \\ \\ = 1 + {sec}^{2} A + {cosec}^{2} A + 2sinA \: sec \: A + 2cosA \: cosec \: A \\ \\ \because \: {sin}^{2} \theta +{cos}^{2} \theta = 1$

Now we know that

$sec \: A = \frac{1}{cos \: A} \\ \\ cosecA = \frac{1}{sinA}$

$1 + \frac{1}{{cos}^{2} A } + \frac{1}{{sin}^{2} A} + \frac{2sinA}{cos \: A} + \frac{2cosA}{sin \: A}$

take LCM of first two trigonometry terms and last two terms

$1 + \frac{ {sin}^{2}A + {cos}^{2}A }{{sin}^{2} A \: {cos}^{2}A } + 2\frac{ {sin}^{2}A + {cos}^{2}A }{sin A\:cos A} \\ \\ = 1 + \frac{1}{{sin}^{2} A \: {cos}^{2}A} + \frac{2}{sinA \: cos \: A} \\ \\ = 1 + {sec}^{2}A \: {cosec}^{2}A+ 2secA \: cosec \: A \\ \\ = ( {1 + secA \: cosecA)}^{2} \\ \\ =R.H.S.$

Hence proved.

Alternative method:

apply these Identity before open square

$sec \: A = \frac{1}{cos \: A} \\ \\ cosecA = \frac{1}{sinA}$

$( {sinA + secA)}^{2} + ( {cosA + cosecA)}^{2} \\ \\=({sinA + \frac{1}{c osA}})^{2} + ( {cosA + \frac{1}{sinA}})^{2}$

$sin^2A+\frac{1}{cos^2A}+\frac{2sinA}{cosA}+cos^2A+\frac{1}{sin^2A}+\frac{2cosA}{sinA}$

take terms of sinA and cos A together

apply identity

$\because \: {sin}^{2} \theta +{cos}^{2} \theta = 1$

$sin^2A+cos^2A+\frac{1}{cos^2A}+\frac{1}{sin^2A}+\frac{2sinA}{cosA}+\frac{2cosA}{sinA}\\\\= 1+\frac{1}{cos^2A}+\frac{1}{sin^2A}+\frac{2sinA}{cosA}+\frac{2cosA}{sinA}$

Take LCM of first two trigonometric terms and last two terms

$1 + \frac{ {sin}^{2}A + {cos}^{2}A }{{sin}^{2} A \: {cos}^{2}A } + 2\frac{ {sin}^{2}A + {cos}^{2}A }{sin A\:cos A} \\ \\ = 1 + \frac{1}{{sin}^{2} A \: {cos}^{2}A} + \frac{2}{sinA \: cos \: A} \\ \\ = 1 + {sec}^{2}A \: {cosec}^{2}A+ 2secA \: cosec \: A \\ \\ = ( {1 + secA \: cosecA)}^{2} \\ \\ =R.H.S.$

Hence proved.

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