Question

√1+cosA /√1-cosA =cosecA +cotA​

Answer 1

Answer:

√1+cosA/√1-cosA

=(√1+cosA)²/(√1-cosA).(√1+cosA)

=1+cosA/√1-cos²A

=1+cosA/sinA

=cosecA+cotA

Answer 2

Given,

$\sf{\sqrt{\dfrac{1+cosA}{1-cosA}}=cosecA+cotA }$

Taking the LHS

$\sf{\implies LHS=\sqrt{\dfrac{1+cosA}{1-cosA}}}$

$\sf{\implies LHS=\sqrt{\dfrac{1+cosA}{1-cosA}}\times \dfrac{1+cosA}{1+cosA} }$    

$\sf{[Multiplying\ numerator\ and\ denominator\ with\ 1 + cos A ]}$

$\sf{\implies LHS=\sqrt{\dfrac{(1+cosA)^2}{1^2-cos^2A}}}$     $\sf{[\because ( a + b ) ( a - b ) = a^2 - b ^2] }$

$\sf{\implies LHS=\sqrt{\dfrac{(1+cosA)^2}{1-cos^2A}}}$

$\sf{\implies LHS=\sqrt{\dfrac{(1+cosA)^2}{sin^2A}}}$     $\sf{[\because 1-cos^2A=sin^2A] }$

$\sf{\implies LHS=\dfrac{1+cosA}{sinA}}$     $\sf{[ Eliminating \ the \ roots \ with \ the \ squares] }$

$\sf{\implies LHS=\dfrac{1}{sinA}+\dfrac{cosA}{sinA}}$     $\sf{[Splitting \ the \ denominators] }$

$\sf{\implies LHS=cosecA+cotA}$

$\sf{\implies LHS=RHS}$

$\boxed{\bold{\sf{\therefore \sqrt{\dfrac{1+cosA}{1-cosA}}=cosecA+cotA }}}$