Question

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Answer 1

$\large\sf\underline{\underline{To\: Prove-}}$

$\rm\implies\dfrac{tan\:A}{sec\:A-1}+\dfrac{tan\:A}{sec\:A+1}=2\:cosec\:A$

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$\large\sf\underline{\underline{Solution-}}$

$\rm\implies\dfrac{tan\:A}{sec\:A-1}+\dfrac{tan\:A}{sec\:A+1}=2\:cosec\:A$

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◯ Consider LHS

$\rm\implies\dfrac{tan\:A}{sec\:A-1}+\dfrac{tan\:A}{sec\:A+1}$

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◯ Taking LCM

$\rm\implies\dfrac{tan\:A\big(sec\:A+1\big)+tan\:A\big(sec\;A-1\big)}{\big(sec\:A-1\big)\big(sec\:A+1\big)}$

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◯ Solve numerator

$\rm\implies\dfrac{tan\:A.sec\:A+tan\:A+tan\:A.sec\:A-tan\:A}{\big(sec\:A-1\big)\big(sec\:A+1\big)}$

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◯ Simplifying it

$\rm\implies\dfrac{2\:tan\:A.sec\:A}{\big(sec\:A-1\big)\big(sec\:A+1\big)}$

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◯ ( A - B ) ( A + B ) = A²- B²

$\rm\implies\dfrac{2\:tan\:A.sec\:A}{sec^2\:A-1}$

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◯ sec² A - 1 = tan² A

$\rm\implies\dfrac{2\:tan\:A.sec\:A}{tan^2\;A}$

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◯ Cancel out tan A

$\rm\implies\dfrac{2\:sec\:A}{tan\;A}$

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◯ tan A = sec A / cosec A

$\rm\implies\dfrac{2\:sec\:A}{\frac{sec\:A}{cosec\:A}}$

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◯ Simplifying it

$\rm\implies\dfrac{2\:sec\:A.cosec\:A}{sec\:A}$

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◯ Cancel out sec A

$\large\boxed{\purple{\rm\implies 2 cosec\:A}}$

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Hence LHS = RHS !!!