Question

Prove that
CotA - CosA ÷Cos^2A equals to CotA ÷ 1+sinA
Please send answer

Answer 1

Answer:

Step-by-step explanation:

If we assume that A is not multiple of 90 degree than it can be solve easily .

Take L.H.S. and change the equation using formula cotA = (cosecA.cosA)

(cosecA . cosA - cosA)/(cosecA .cosA + cosA)

Take out common term from numerator and denominator and cancel it if our assumption is right.

(cosecA-1)/(cosecA+1) which is equal to R.H.S. of given equation.

Hence proved.

hope it helps!

Answer 2

To prove :-

$\frac{cot \: a - cos \: a}{ {cos}^{2}a } = \frac{cot \: a}{1 + sin \: a}$

Proof :-

As cosec a×Cos a is cot a,

Therefore,

L. H. S,

$= > \frac{cosec \: a.cos \: a - cos \: a}{ {cos}^{2} a}$

$= > \frac{cos \: a(cosec \: a - 1)}{ {cos}^{2}a }$

$= > \frac{cos \: a( \frac{1}{sin \: a} - 1)}{1 - {sin}^{2}a }$

$= > \frac{cos \: a( \frac{1 - sin \: a}{sin \: a})}{(1 - sin \: a)(1 + sin \: a)}$

$= > cos \: a \times \frac{1 - sin \: a}{sin \: a} \times \frac{1}{(1 - sin \: a)(1 + sin \: a)}$

$= > \frac{cos \: a}{sin \: a} \times \frac{1}{1 + sin \: a}$

$= >\frac{cot \: a}{1 + sin \: a} = \: R. H. S$

As L. H. S=R. H. S

Hence proved.