Question

If cosecA - cotA = q , then show that q 2 -1/q 2 +1 +cosA=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

Step-by-step explanation:

CosecA-CotA= q---(1)

Taking reciprocal,

1/CosecA-CotA=1/q

1(CosecA+CotA)/CosecA-Cot A(CosecA+CotA)=1/q

CosecA+CotA/Cosec^2A-Cot^2A=1/q

CosecA+CotA/1= 1/q {∵Cosec^2A-Cot^2A=1} ---(2)

(1)+(2),

2CosecA=q+(1/q)=q^2+1/q----(3)

(1)-(2),

-2CotA=q-(1/q)=q^2-1/q----(4)

(4)/(3),

-2CotA/2CosecA=q^2-1/q÷q^2+1/q

-CotA/CosecA=q^2-1/q^2+1

(-CosA/SinA)÷(1/SinA)=q^2-1/q^2+1 {∵CotA=CosA/SinA & CosecA=1/SinA}

(-CosA/SinA)(SinA)=q^2-1/q^2+1

-CosA=q^2-1/q^2+1-----(5)

Now, LHS

q^2-1/q^2+1+CosA

⇒ -CosA+CosA=0(from (5) ),RHS.

Hence, proved.

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