Math, asked by ditya329, 3 days ago

solve this problem.... don't spam​

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Answered by amansharma264
0

EXPLANATION.

\sf \displaystyle (tan \theta + sec \theta )^{2} + ( tan \theta - sec \theta )^{2} = \frac{1 + sin^{2} \theta}{1 - sin^{2} \theta} + \frac{1 + sin^{2} \theta }{cos^{2} \theta}

As we know that,

Fundamental trigonometric identities.

⇒ sin²θ + cosθ = 1.

⇒ 1 + tan²θ = sec²θ.

⇒ 1 + cot²θ = cosec²θ.

⇒ secθ = 1/cosθ.

⇒ tanθ = sinθ/cosθ

Using this identities in this question, we get.

First we solve LH.S.

\sf \displaystyle (tan \theta + sec \theta )^{2} + ( tan \theta - sec \theta )^{2}

\sf \displaystyle ( tan^{2} \theta + sec^{2} \theta + 2 tan \theta sec \theta ) + ( tan^{2} \theta + sec^{2} \theta - 2 tan \theta sec \theta )

\sf \displaystyle  tan^{2} \theta + sec^{2} \theta + 2 tan \theta sec \theta  +  tan^{2} \theta + sec^{2} \theta - 2 tan \theta sec \theta

\sf \displaystyle  2tan^{2} \theta + 2sec^{2} \theta

\sf \displaystyle  2 (sec^{2} \theta - 1) + 2sec^{2} \theta

\sf \displaystyle 2sec^{2} \theta - 2 + 2 sec^{2} \theta

\sf \displaystyle 4 sec^{2} \theta - 2

Now, we solve R.H.S.

\sf \displaystyle \frac{1 + sin^{2} \theta}{1 - sin^{2} \theta} + \frac{1 + sin^{2}\theta }{cos^{2}\theta }

\sf \displaystyle \frac{1 + (1 - cos^{2} \theta)}{cos^{2}\theta } + \frac{1 + (1 - cos^{2}\theta) }{cos^{2}\theta }

\sf \displaystyle \frac{1 + 1 - cos^{2}\theta }{cos^{2} \theta} + \frac{1 + 1 - cos^{2} \theta}{cos^{2} \theta}

\sf \displaystyle \frac{2 - cos^{2}\theta }{cos^{2} \theta} + \frac{2 - cos^{2} \theta}{cos^{2} \theta}

\sf \displaystyle \frac{2 - cos^{2}\theta + 2 - cos^{2} \theta }{cos^{2} \theta}

\sf \displaystyle \frac{4 - 2 cos^{2}  \theta}{cos^{2} \theta}

\sf \displaystyle \frac{4}{cos^{2} \theta} - \frac{2 cos^{2}\theta }{cos^{2} \theta}

\sf \displaystyle 4 sec^{2} \theta - 2

Hence proved.

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