Question

Prove that √((cosec^ A+1)/(cosec^ A-1)) = sec A + tan A

Answer 1

Answer:

Hey there !!

Prove that :-

\sqrt{ \frac{cosec \theta - 1}{ cosec \theta + 1} } + \sqrt{ \frac{cosec \theta + 1}{cosec \theta - 1} } = 2 \sec \theta. \\ \\

Solution :-

Solving LHS :-

= \sqrt{ \frac{cosec \theta - 1}{ cosec \theta + 1} } + \sqrt{ \frac{cosec \theta + 1}{cosec \theta - 1} } \\ \\ = \sqrt{ \frac{cosec \theta - 1}{ cosec \theta + 1} \times \frac{ \cosec \theta - 1}{\cosec \theta - 1} } + \sqrt{ \frac{cosec \theta + 1}{cosec \theta - 1} \times \frac{\cosec \theta + 1}{\cosec \theta + 1} } . \\ \\ = \sqrt{ \frac{ {(\cosec \theta - 1)}^{2} }{ {\cosec}^{2} \theta - {1}^{2} } } + \sqrt{ \frac{ {(\cosec \theta + 1)}^{2} }{ {\cosec}^{2} \theta - {1}^{2} } } . \\ \\ = \sqrt{ \frac{ {(\cosec \theta - 1)}^{2} }{ { - \cot}^{2} \theta} } + \sqrt{ \frac{ {(\cosec \theta + 1)}^{2} }{ { - \cot}^{2} \theta} } . \\ \\ = \frac{ \cosec\theta - 1}{ \cot\theta} + \frac{ \cosec\theta + 1}{ \cot\theta} . \\ \\ = \frac{ \cosec\theta }{ \cot\theta} - \cancel\frac{1}{ \cot\theta} + \frac{ \cosec\theta}{ \cot\theta} + \cancel\frac{1}{ \cot\theta} . \\ \\ = \frac{ \frac{1}{ \cancel{\sin\theta} }}{ \frac{ \cos\theta}{ \cancel{\sin\theta} }} + \frac{ \frac{1}{ \cancel{\sin\theta} }}{ \frac{ \cos\theta}{ \cancel{\sin\theta} }} . \\ \\ = \frac{1}{ \cos\theta} + \frac{1}{ \cos \theta} . \\ \\ = \sec \theta + \sec \theta. \\ \\ = \boxed{ \boxed{ \pink{ = 2 \sec \theta.}}} \checkmark \checkmark.

•°• LHS = RHS .

✔✔ Hence, it is proved ✅✅.

THANKS

Answer 2

Answer:

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Step-by-step explanation:

We need to prove that ,

 √((cosec^ A+1)/(cosec^ A-1)) = sec A + tan A

(cosecA+1)\(cosecA−1)=(secA+tanA)∧2

L.H.S

cosec A = 1 / sin A

cosecA+1=>1/sinA+1

=> (1+sinA)/sinA−−−(i)

cosecA−1=>1/sinA−1

=> (1−sinA)/sinA−−−(ii)

Divide (i) and (ii)

[(1+sinA)/sinA]/[(1−sinA)/sinA]

(1+sinA)/(1−sinA)

Multiplying by ( 1+sin A ) in numerator and denominator

[(1+sinA)2]/(1−sin2A)

1 - sin^2 A = cos ^ 2 A = ( cos A ) ^ 2

(1+sinA)2/(cosA)2

a^m ÷ b^m = ( a ÷ b ) ^m

(1+sinA/cosA)2

1 / cos A = sec A

sin A / cos A = tan A

(secA+tanA)2

L.H.S = R.H.S