Question

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


Need prove :(
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Answer 1

Step-by-step explanation:

$\huge\boxed{\fcolorbox{cyan}{apricot}{⛄~MysticalBlush~⛄}}$

$\huge\bold\red{LHS}$ :-

$\huge\bold\green{Formula \: used}$ ,

$\frac{tan \: a \: (sec \: a \: + 1) \: + \: tan \: a \: (sec \: a \: - 1)}{(sec \: a \: - \: 1) \: (sec \: a \: + \: 1)} \\ \\ = > \frac{tan \: a \: (sec \: a \: + \: 1 \: + \: sec \: a \: - \: 1)}{ {sec}^{2} \: - \: 1} \\ \\ = > \frac{tan \: a \: (2 \: sec \: a)}{ {sec}^{2} \: - \: 1} \\ \\ = > \frac{2 \: sec \: a \: \times \: tan \: a }{1 \: + \: {tan}^{2} \: - \: 1 } \\ \\ = > \frac{2 \: sec \: a \: \: tan \: a}{ {tan}^{2} \: a } \\ \\ = > \frac{2 \: sec \: a}{tan \: a } \\ \\ = > 2 \: \: sec \: a \: \: cot \: a \\ \\ = > 2 \: \times \: \frac{1}{cos \: a} \times \frac{cos \: a}{sin \: a} \\ \\ = > 2 \: \times \: cosec \: a \:$

$\huge\bold\orange{Hence \: proved}$ $\huge\bold\green{~✔~}$

Answer 2

Answer:-

To Prove:

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

Taking LCM we get,

$\implies \sf \: \frac{ \tan(a)( \sec(a) + 1) + \tan(a) ( \sec(a) - 1)}{( \sec(a) - 1)( \sec(a) + 1) } = 2 \csc(a) \\\\ \\ \implies \sf \: \frac{ \tan(a) \sec(a) + \tan(a) \sec(a) - \tan(a) }{ { \sec}^{2} (a) - {1}^{2} } = 2 \: \csc(a)$

[ (a + b)(a - b) = a² - b² ]

Using the identity sec² a - 1 = tan² a in LHS we get,

$\: \implies \sf \: \frac{2 \tan(a) \sec(a) }{ { \tan}^{2} (a)} = 2 \: \csc(a)$

using tan a = sin a/cos a and sec a = 1/cos a in LHS we get,

$\implies \sf \: 2 \times \frac{ \frac{1}{ \cos(a) } }{ \frac{ \sin(a) }{ \cos(a) } } = 2 \: \csc(a) \\ \\ \\ \implies \sf \: 2 \times \frac{1}{ \cos(a) } \times \frac{ \cos(a) }{ \sin(a) } = 2 \: \csc(a) \\ \\ \\ \implies \sf \: \frac{2}{ \sin(a) } = 2 \csc(a)$

using 1/sin a = Cosec a we get,

⟶ 2 Cosec a = 2 Cosec a.

Hence Proved.