Question

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

Answer 1

There is a mistake in the question. It should be (q²-1)/(q²+1) -1/cos A = 0

Cosec A - cot A = q

1/q = 1/(cosec A - cot A)

      = (cosecA + cotA) / [(cosec A + cotA) (cosec A  - cot A)

      = (cosec A + cot A )/ 1

q + 1/q =  (q²+1) / q =  2 cosec A    -- (1)

q - 1/q = (q² - 1) / q = - 2 cot A    -- (2)

(2) ÷ (1)  =>  

  (q² -1) / (q² + 1)   =  - cosec A / cot A

                              = - sec A 

LHS =  (q² - 1) / (q² + 1)  + 1/cos A

       =  -sec A + sec A

       = 0

Answer 2

Answer:

Important Formulas:

(1). 2sin²(A/2) = 1 - cosA

(2). cos²(A/2) - sin²(A/2) = cosA

Step-by-step explanation:

cosecA - cotA = q.

$\Rightarrow \frac{1}{sinA} - \frac{cosA}{sinA} = q$

$\Rightarrow \frac{2sin^2A}{sinA} = q$

$\Rightarrow \frac{2sin^2\frac{A}{2} }{2sin\frac{A}{2}cos\frac{A}{2}} = q$

$\Rightarrow \frac{sin\frac{A}{2}}{cos\frac{A}{2}} = q$

$\Rightarrow q = tan\frac{A}{2}$

Now,

$\frac{q^2 - 1}{q^2 + 1} + cosA$

$\Rightarrow  \frac{tan^2\frac{A}{2}  - 1}{tan^2\frac{A}{2}  + 1} + cosA$

$\Rightarrow \frac{\frac{sin^2\frac{A}{2}}{cos^2\frac{A}{2}}-1}{\frac{sin^2\frac{A}{2}}{cos^2\frac{A}{2} } + 1} + cosA$

$\Rightarrow \frac{sin^2\frac{A}{2} - cos^2\frac{A}{2}}{sin^2\frac{A}{2} + cos^2\frac{A}{2}} + cosA$

$\Rightarrow -cosA + cosA$

$\bold 0$

Hope it helps!