Question

prove(1+cotA-cosecA) (1+tanA+secA)=2​

Answer 1

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

$\tt Prove:$

$\bf \big(1+cotA-cosecA\big) \big(1+tanA+secA\big)=2$

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

$\sf \bf LHS$

$\sf \bf \implies \Big(1+cot\:A-cosec\:A\Big) \Big(1+tan\:A+sec\:A\Big)$

$\sf \bf \implies \Bigg(1+\dfrac{cos\:A}{sin\:A} - \dfrac{1}{sin\:A}\Bigg) \Bigg(1+\dfrac{sin\:A}{cos\:A} +\dfrac{1}{cos\:A}\Bigg)$

$\sf \bf \implies \Bigg(\dfrac{sin\:A+cos\:A-1}{sin\:A}\Bigg)\Bigg(\dfrac{cos\:A+sin\:A+1}{cos\:A} \Bigg)$

$\sf \bf \implies \dfrac{{\Big(sin\:A+cos\:A\Big)}^{2}-{\Big(1\Big)}^{2}}{sin\:A \times cos\:A}$

$\sf \bf \implies \dfrac{{sin}^{2}\:A+{cos}^{2}\:A+2\big(sin\:A\big) \big(cos\:A\big) - 1}{sin\:A\:cos\:A}$

$\sf \bf \implies \dfrac{1+2sin\:A\:cos\:A- 1}{sin\:A\:cos\:A}$

$\sf \bf \implies \dfrac{\cancel {1} +2sin\:A\:cos\:A\cancel {- 1}}{sin\:A\:cos\:A}$

$\sf \bf \implies \dfrac{2sin\:A\:cos\:A}{sin\:A\:cos\:A}$

$\sf \bf \implies \dfrac{2\cancel {sin\:A\:cos\:A}}{\cancel{sin\:A\:cos\:A}}$

$\implies {\boxed {\boxed {\color {blue} {\sf \bf 2}}}}$

$\sf \bf RHS$

${\color {purple} {\sf Used\:Identities}}$

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

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

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

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

${\color {green} {\tt Identities}}$

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

$\sf \bf {(a-b)}^{2}={a}^{2}-2ab+{b}^{2}$

$\sf \bf (a+b) (a-b) ={a}^{2} - {b}^{2}$

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

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

$\sf \bf 1+{cot}^{2}\:A ={cosec}^{2}\:A$