Question

prove (sec A + tan A)*2 = cosec A + 1/cosec A - 1​

Answer 1

Step-by-step explanation:

LHS :

(secA + tanA)²

( 1/cosA + sinA/cosA)²

(1 + sinA)² / cos²A

we know that ,

cos²A = 1 - sin²A

(1+sinA) (1 + sinA) / {1 - sin²A}

Also ,

1 - sin²A = (1+sinA) (1 - sinA) [ ∵ a² - b² = (a+b) (a - b) ]

(1+sinA) (1 + sinA) / (1+sinA) (1 - sinA)

(1 + sinA ) gets cancelled on numerator and denominator ,

=> 1 + sinA / 1 - sinA

sinA = 1/cosecA

=> 1 + ( 1/cosecA) / [ 1 - (1/cosecA) ]

=> \frac{ \frac{cosecA + 1}{cosecA} }{ \frac{cosecA - 1}{cosecA} }

cosecA

cosecA−1

cosecA

cosecA+1

=> cosec A gets cancelled ,on both sides ,

hence ,

LHS becomes ,

cosecA + 1/cosecA - 1

Answer 2

To Prove :-

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

Prove :-

$\sf We \: have$,

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

Take L.H.S :-

$\implies \sf (sec \: A + tan \: A)^{2}$

• First convert sec A and tan A in terms of cos A and sin A. So, sec A = 1/cos A and tan A = sin A/cos A.

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

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

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

• cos² A = 1 - sin² A

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

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

• Use a² - b² = (a + b)(a - b) in denominator.

$\implies \sf \dfrac{(1 + sin \: A) \times \cancel{(1 + sin \: A)}}{(1 - sin \: A) \cancel{(1 + sin \: A)}}$

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

• sin A = 1/cosec A

$\implies \sf \dfrac{1 + \dfrac{1}{cosec \: A}}{1 - \dfrac{1}{cosec \: A}}$

$\implies \sf \dfrac{\dfrac{cosec\: A + 1}{\cancel{cosec \: A}}}{\dfrac{cosec\: A - 1}{\cancel{cosec \: A}}}$

$\implies \sf \dfrac{cosec \: A + 1}{cosec \: A - 1}$

$\implies \sf R.H.S$

Hence, Proved!!