Question

please solve this question​

Answer 1

Answer:-

We have to prove:-

$\sf \: \dfrac{ \tan A }{ \sec A - 1} + \dfrac{ \tan A}{ \sec A + 1} = 2 \csc A$

Taking LCM in LHS we get,

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

Taking tan A common in numerator and using (a + b)(a - b) = a² - b² in LHS we get,

$\sf \implies \: \dfrac{ \tan A( \sec A - \cancel{ 1 }+ \sec A+ \cancel{ 1}) }{ { \sec }^{2} A - {1}^{2} } = 2 \csc A \\ \\ \\ \sf \implies \: \frac{ \tan A \times 2 \sec A}{{ \sec }^{2} A - {1}} = 2 \csc A$

using tan² A = sec² A - 1 we get,

$\implies \sf \: \dfrac{ \tan A \times 2 \sec A }{ \tan ^{2} A } = 2 \csc A \\ \\ \\ \implies \sf \: \frac{2 \sec A }{ \tan A } = 2 \csc A$

Using tan A = sin A/cos A and sec A = 1/cos A we get,

$\: \implies \sf \: \dfrac{2 \times \frac{1}{ \cancel{ \cos A }} }{ \frac{ \sin A }{ \cancel{\cos A }} } = 2 \csc A \\ \\ \\ \implies \sf \:2 \times \frac{1}{ \sin A } = 2 \csc A$

Using cosec A = 1/sin A in LHS we get,

$\: \implies \sf \:2 \csc A = 2 \csc A$

Hence, Proved.

Answer 2

To Prove:-

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

Identities used:-

• a² - b² = (a + b) (a - b)

• tan² A = sec² A - 1

• sec A = $\dfrac{1}{cos\:A}$

• tan A = $\dfrac{sin\:A}{cos\:A}$

• cosec A = $\dfrac{1}{sin\:A}$

Proof:-

⇒ L.H.S = $\dfrac{tan\:A}{sec\:A-1}+\dfrac{tan\:A}{sec\:A+1}$

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= $\dfrac{tan\:A(sec\:A+1)+tan\:A(sec\:A-1)}{(sec\:A-1)(sec\:A+1)}$

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= $\dfrac{tan\:A(sec\:A+1)(sec\:A-1)}{(sec\:A-1)(sec\:A+1)}$

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${\underline{Now, \:using \:identity \: a² - b² = (a + b) (a - b)}}$

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= $\dfrac{tan\:A(sec\:A-1+sec\:A+1)}{(sec²\:A-1²)}$

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= $\dfrac{tan\:A × 2 \: sec\: A}{(sec²\:A-1)}$

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${\underline{Using\: identity, \:tan² \:A = sec² \:A - 1}}$

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= $\dfrac{tan\:A × 2 \: sec\: A}{tan² \:A}$

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= $\dfrac{2\:sec\: A}{tan \:A}$

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${\underline{Now, \:using \:identity \:sec\: A=\dfrac{1}{cos\:A}}}$

${\underline{and\: also \:tan \:A\: =\dfrac{sin\:A}{cos\:A}}}$

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= $2×\dfrac{1}{sin\:A}$

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${\underline{Using \:identity\:cosec\:A\:= \:\dfrac{1}{sin\:A}}}$

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= ${\bold{2\:cosec\:A\:=\:R.H.S}}$

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$\text{\boxed{\orange{L.H.S = R.H.S}}}$

HENCE Proved.

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