Question

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Answer 1

TO PROVE :–

$\\ \implies\rm \sqrt\dfrac{1 + \cos(A) }{1 - \cos(A) } = \dfrac{1 + \cos(A) }{ \sin(A) }$

SOLUTION :–

• Let's take L.H.S. –

$\\ \rm \: \: = \sqrt\dfrac{1 + \cos(A) }{1 - \cos(A) } \: \:$

• Now Rationalization of denominator –

$\\ \rm \: \: = {\sqrt{\dfrac{1 + \cos(A) }{1 - \cos(A) } \times\dfrac{1 + \cos(A) }{1 + \cos(A)}} }\: \:$

$\\ \rm \: \: = {\sqrt{\dfrac{ \{1 + \cos(A)\}^{2} }{ \{1 - \cos(A) \} \{ 1 + \cos(A)\}}}}$

• We know that –

$\\ \large\longrightarrow{ \boxed{ \rm(a - b)(a + b) = {a}^{2} - {b}^{2} }}$

• So that –

$\\ \rm \: \: = {\sqrt{\dfrac{ \{1 + \cos(A)\}^{2} }{ \{1 - \cos^{2} (A) \}}}}$

• We also know that –

$\\ \large\bf\implies \sin^{2}( \theta) +\cos^{2}( \theta) = 1$

$\\ \large\bf\implies \sin^{2}( \theta)= 1 - \cos^{2}( \theta)$

$\\ \large\bf\implies 1 - \cos^{2}( \theta) = \sin^{2}( \theta)$

• So that –

$\\ \rm \: \: = \:\: {\sqrt{\dfrac{ \{1 + \cos(A)\}^{2} }{\sin^{2} (A)}}}$

$\\ \rm \: \: = \: \: \dfrac{ 1 + \cos(A)}{\sin(A)}$

$\\ \rm \: \: = \: \: R.H.S.$

$\\ \large \longrightarrow{ \boxed{ \rm Hence \: \:Proved }}$

Answer 2

TO PROVE :–

$\\ \implies\rm \sqrt\dfrac{1 + \cos(A) }{1 - \cos(A) } = \dfrac{1 + \cos(A) }{ \sin(A) }$

SOLUTION :–

• Let's take L.H.S. –

$\\ \rm \: \: = \sqrt\dfrac{1 + \cos(A) }{1 - \cos(A) } \: \:$

• Now Rationalization of denominator –

$\\ \rm \: \: = {\sqrt{\dfrac{1 + \cos(A) }{1 - \cos(A) } \times\dfrac{1 + \cos(A) }{1 + \cos(A)}} }\: \:$

$\\ \rm \: \: = {\sqrt{\dfrac{ \{1 + \cos(A)\}^{2} }{ \{1 - \cos(A) \} \{ 1 + \cos(A)\}}}}$

• We know that –

$\\ \large\longrightarrow{ \boxed{ \rm(a - b)(a + b) = {a}^{2} - {b}^{2} }}$

• So that –

$\\ \rm \: \: = {\sqrt{\dfrac{ \{1 + \cos(A)\}^{2} }{ \{1 - \cos^{2} (A) \}}}}$

• We also know that –

$\\ \large\bf\implies \sin^{2}( \theta) +\cos^{2}( \theta) = 1$

$\\ \large\bf\implies \sin^{2}( \theta)= 1 - \cos^{2}( \theta)$

$\\ \large\bf\implies 1 - \cos^{2}( \theta) = \sin^{2}( \theta)$

• So that –

$\\ \rm \: \: = \:\: {\sqrt{\dfrac{ \{1 + \cos(A)\}^{2} }{\sin^{2} (A)}}}$

$\\ \rm \: \: = \: \: \dfrac{ 1 + \cos(A)}{\sin(A)}$

$\\ \rm \: \: = \: \: R.H.S.$

$\\ \large \longrightarrow{ \boxed{ \rm Hence \: \:Proved }}$