Question

Is cosecA + cotA = p then prove that cosA = p^2 - 1 / p^2 + 1

Answer 1

Cos A = (p²–1)÷(p²+1)

Answer 2

Solution:

We have,

cosecA+cotA = p -----(1)

Now, cosec²A-cot²A=1

=> (cosecA+cotA)(cosecA-cotA)=1

=> p×(cosecA-cotA)=1

=> cosecA-cotA = $\frac{1}{p}$----(2)

adding and subtracting (1) and

(2), we get

i)(cosecA+cotA)+(cosecA-cotA)=$p+\frac{1}{p}$

=> 2cosecA= $\frac{(p^{2}+1)}{p}$

=>cosecA= $\frac{(p^{2}+1)}{2p}$ ---(3)

ii)(cosecA+cotA)-(cosecA-cotA)=$p-\frac{1}{p}$

=> 2cotA= $\frac{(p^{2}-1)}{p}$

=>cotA= $\frac{(p^{2}-1)}{2p}$ ---(4)

Now ,

cosA = $\frac{cotA}{cosecA}$

/* from (3) and (4) */

=$\frac{\frac{(p^{2}-1)}{2p}}{\frac{(p^{2}+1)}{2p}}$

= $\frac{p^{2}-1}{p^{2}+1}$

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