Question

if cosec A- cot A = q, show that q square-1/ q square+1 + cos A =0

Answer 1

refer to the attached image for proof....

Answer 2

q square-1/ q square+1 + cos A =0

Hence proved below.

Step-by-step explanation:

Given:

We have,      

q = cosec A − cot A

$q^2 = (cosec A - cot A)^2$

$q^2 = cosec^2A + cot^2A - 2 cosec A. cot A$

$q^2 = \frac{1}{sin^2A} + \frac{cos^2A}{sin^2A}- 2\frac{cos A}{sin^2 A}$

$q^2 =\frac{1 + cos^2A - 2 cos A}{sin^2A}$

$q^2 =\frac{(1 - cos A)^2}{ 1 - cos^2A}$

$q^2 =\frac{ (1-cos A)(1 - cos A)}{(1 - cos A)(1 + cos A)}$

$q^2=\frac{ (1-cos A)}{(1 + cos A)}$             [1]

Now,

LHS = $\frac{q^2-1}{ q^2+1} + cos A$

       = $\frac{\frac{1-cosA}{1+cosA}-1 } {\frac{1-cosA}{1+cosA}+1}+ cos A$            [from equation 1]

       = $\frac{\frac{(1-cosA-(1 + cos A))}{(1 + cos A)} }{\frac{1-cosA+1 + cos A}{1+cosA} }+cosA$

       = $\frac{ 1 - cos A - 1 - cos A}{1 - cos A + 1 + cos A} + cos A$

       = $\frac{-2 cos A}{2} + cos A$

       = - cos A + cos A

       = 0

       = RHS

Hence proved.