Question

tan square A/tan square A-1+Cosec square A/sec square A-Cosec square A=1/1-2 cos square A

Answer 1

Answer:

We have proved that L.H.S=R.H.S.

i.e;

$\dfrac{tan^2A}{tan^2A-1}+\dfrac{cosec^2A}{sec^2A-cosec^2A}=\dfrac{1}{1-2cos^2A}$

Step-by-step explanation:

Here, we are evaluating by the given equation.

L.H.S

$\dfrac{tan^2A}{tan^2A-1}+\dfrac{cosec^2A}{sec^2A-cosec^2A}$

Here we are using some trigonometric formula as given below:

$tanx=\dfrac{sinx}{cosx}\\cosecx=\dfrac{1}{sinx}\\secx=\dfrac{1}{cosx}$

So the equation will be:

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

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

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

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

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

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

Again we know the formula

i.e;

$sin^2x+cos^2x=1$

Or,

$sin^2x=1-cos^2x$

Substituting this in the above equation:

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

=R.H.S

Answer 2

Answer:

Step-by-step explanation: