Math, asked by Anonymous, 4 months ago

Answer the above attachment question....✌️​

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Answered by suraj5070
205

\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}}}

_________________________________________

\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

\sf {\mathbb{\colorbox {orange} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {lime} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {aqua} {@suraj5070}}}}}}}}}}}}}}}


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