Question

Answer the above attachment question....✌️​

Answer 1

$\sf \huge {\boxed {\mathbb {QUESTION}}}$

$\sf Prove \:that$

$\sf (tan\: \theta + sec\: \theta-1) (tan\: \theta + sec\: \theta+1)= \dfrac{2 sin \: \theta}{1-sin \: \theta}$

$\sf\huge {\boxed {\mathbb {ANSWER}}}$

$\sf \implies (tan\: \theta + sec\: \theta-1) (tan\: \theta + sec\: \theta+1)= \dfrac{2 sin \: \theta}{1-sin \: \theta}$

$\sf LHS$

$\sf\implies (tan\: \theta + sec\: \theta-1) (tan\: \theta + sec\: \theta+1)$

$\sf \implies {(tan\: \theta +sec\: \theta)}^{2}-{(1)}^{2}$

$\sf \implies {tan}^{2}\: \theta +2 tan\: \theta.sec\: \theta + {sec}^{2}\: \theta - 1$

$\sf \implies {tan}^{2}\: \theta +2 tan\: \theta.sec\: \theta \cancel {+1} + {tan}^{2}\: \theta \cancel {- 1}$

$\sf \implies 2 {tan}^{2}\: \theta +2 tan\: \theta.sec\: \theta$

$\sf \implies 2 tan\: \theta (tan\: \theta +sec\: \theta)$

$\sf\implies 2 \frac{sin\: \theta}{cos\: \theta}(\dfrac{sin\: \theta}{cos\: \theta} + \dfrac{1}{cos\: \theta})$

$\sf\implies 2 \dfrac{sin\: \theta}{cos\: \theta}(\dfrac{sin\: \theta+1}{cos\: \theta})$

$\sf\implies \dfrac{2 sin\: \theta (1+sin\: \theta)}{{cos}^{2}\: \theta}$

$\sf\implies \dfrac{2 sin\: \theta (1+sin\: \theta)}{1-{sin}^{2}\: \theta}$

$\sf \implies \dfrac{2 sin\: \theta \cancel {(1+sin\: \theta)}}{(1-sin\: \theta) \cancel {(1+sin\: \theta)}}$

$\sf\implies \dfrac{2 sin \: \theta}{1-sin \: \theta}$

$\sf RHS$

$\sf\huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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$\sf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

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

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

$\sf1+{cot}^{2}\: \theta={cosec}^{2}\: \theta$

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