Question

cosecA + cotA= K then prove that cos A = k2-1/K2+1

Answer 1

1/sinA+cosA/sinA=k
1+cosA/sinA=k
1+cos^2A/sin^2A= k^2
componendo dividendo

1+cos^2A-sin^2A/sin^2A+sin^2A= k^2-1/k^2+1
2cos^2A/2sun^2A=k2-1/k2+1

cot2A

Answer 2

$\cos A=\dfrac{k^2-1}{k^2+1}$, proved.

Step-by-step explanation:

We have,

$\csc A +\cot A$ = K

To prove that, $\cos A=\dfrac{k^2-1}{k^2+1}$.

∴ $\csc A +\cot A$ = K

Using the trigonometric identity,

$\csc A$ = $\dfrac{1}{\sin A}$ and

$\cot A$ = $\dfrac{\cos A}{\sin A}$

$\dfrac{1}{\sin A} +\dfrac{\cos A}{\sin A}$ = K

⇒ $\dfrac{1+\cos A}{\sin A}$ = K

R.H.S. = $\dfrac{k^2-1}{k^2+1}$

Put k =  $\dfrac{1+\cos A}{\sin A}$, we get

= $\dfrac{(\dfrac{1+\cos A}{\sin A})^2-1}{(\dfrac{1+\cos A}{\sin A})^2+1}$

= $\dfrac{(1+\cos A)^2-\sin^2 A}{(1+\cos A)^2+\sin^2 A}$

Using the algebraic identity,

$(a+b)^{2}$ = $a^{2}$ + 2ab + $b^{2}$

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

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

Using the trigonometric identity,

$\sin^2 A+\cos^2 A$= 1

⇒ $\cos^2 A$ = 1 - $\sin^2 A$

= $\dfrac{\cos^2 A+\cos^2 A+2\cos A}{1+1+2\cos A}$

= $\dfrac{2\cos^2 A+2\cos A}{2+2\cos A}$

= $\dfrac{2\cos A(1+\cos A)}{2(1+\cos A)}$

= $\cos A$

= R.H.S. , proved.

Thus, $\cos A=\dfrac{k^2-1}{k^2+1}$, proved.