Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Using trigonometric formulas,

$\cos e c^{2} A - c o t^{2} A = 1$

$\cos e c^{2} A = 1 + c o t^{2} A$

Using $\cos e c A = \frac{1}{\sin A}$ .

$\Rightarrow \frac{1}{\sin^{2} A} = 1 + c o t^{2} A$

Rearranging further we get,

$\Rightarrow \sin^{2} A = \frac{1}{1 + c o t^{2} A}$

Now, take square roots on both sides, we get

$\Rightarrow \sin A = \pm \frac{1}{\sqrt{1 + c o t^{2} A}}$

Hence we get the required result for sine function.

Now using the result,

$\sin^{2} A = \frac{1}{1 + c o t^{2} A}$

As we know, $\Rightarrow \sin^{2} A = 1 - \cos^{2} A$

$\Rightarrow 1 - \cos^{2} A = \frac{1}{1 + c o t^{2} A}$

Rearranging further we get,

$\cos^{2} A = 1 - \frac{1}{1 + c o t^{2} A}$

$\Rightarrow \cos^{2} A = \frac{1 + c o t^{2} A - 1}{1 + c o t^{2} A} = \frac{c o t^{2} A}{1 + c o t^{2} A}$

Using, $C o s A = \frac{1}{s e c A}$

$\Rightarrow \frac{1}{s e c^{2} A} = \frac{c o t^{2} A}{1 + c o t^{2} A}$

$\Rightarrow s e c^{2} A = \frac{1 + c o t^{2} A}{c o t^{2} A}$

Now, square roots on both sides, we get

$s e c A = \pm \sqrt{\frac{1 + c o t^{2} A}{c o t^{2} A}} = \pm \frac{\sqrt{1 + c o t^{2} A}}{c o t A}$

Hence we get the required result for secant function.

Now, as we know $\tan A = \frac{1}{c o t A}$

Hence we get the required result for tangent function.

Watch 3-min video & get full concept clarity