Question

sin2theta/tan2theta=1+ cot2theta -cos2theta​

Answer 1

Answer:

your answer

i hope half you

Answer 2

Answer:

HOPE IT HELPS YOU

Step-by-step explanation:

$\frac{sec^2-sin^2}{tan^2} = 1 +cot^2-cos^2$

Solving R.H.S

$\frac{sec^2-sin^2}{tan^2} = 1 +cot^2-cos^2\\\\\\\frac{sec^2-sin^2}{tan^2} = cosec^2-cos^2\\\\\frac{sec^2-sin^2}{tan^2} = \frac{1}{sin^2}-cos^2 \\\\\frac{sec^2-sin^2}{tan^2}= \frac{1-sin^2cos^2}{sin^2} \\\\\\solving L.H.S\\\\sec^2 = 1/cos^2 and tan^2= sin^2/cos^2\\\\\frac{\frac{1}{cos^2}-sin^2 }{\frac{sin^2}{cos^2} } \\\\\\\frac{\frac{1-sin^2.cos^2}{cos^2} }{\frac{sin^2}{cos^2} } \\\\\frac{1-sin^2.cos^2}{sin^2} }$L.H.S = R.H.S