Question

1+tan²theta+1+1/tan theta=1/sin²theta-sin²theta​

Answer 1

Solution :

1 + tan² theta + 1 + 1/tan theta

>> 1 + sin²theta/cos²theta + 1 + cos theta/sin theta

>> 2 + (sin²theta/cos²theta) + ( cos theta/sin theta)

>> (1 + sin²theta/cos²theta) + ( 1 + cos theta/sin theta)

>> 1 + (sin²theta + cos²theta/cos²theta) + cos theta/sin theta

>> 1 + (1/cos²theta) + cos theta/sin theta

Taking LCM

>> [ cos²theta sin theta + sin theta + cos³theta]/[ sin theta cos² theta ]

>> [ sin theta {2cos²theta + sin ²theta} + cos³theta]/[sin theta cos² theta]

>> [ sin ³theta + 2cos ²theta sin theta + cos ³theta]/[ sin theta cos² theta]

>> [ (sin theta + cos theta)( sin²theta + cos ²theta - sin theta cos theta) + 2 cos²theta sin theta ]/[ sin theta cos² theta]

>> [ (sin theta + cos theta)( 1 - sin theta cos theta) + 2cos²theta sin theta ]/[ sin theta cos² theta]

>> [ sin theta - sin²theta cos theta + cos theta + sin theta cos²theta ]/[ sin theta cos² theta]

Cancelling sin theta from all sides

>> (1 - sin theta cos theta + tan theta + cos²theta)/cos²theta

>> sec² theta - tan theta + sin theta/(cos³theta) + 1

Multiplying and dividing by sin²theta

>> [ sec ²theta sin ²theta - tan theta sin²theta + sin ³theta/cos³theta ]/sin²theta

>> [ tan²theta - sin³theta/costheta + tan³theta ]/sin² theta

>> [ sin²theta/cos²theta - sin³theta/cos theta + sin³theta/coa³theta]/sin²theta

>> [ 1/cos³theta( sin²theta cos theta - sin³theta cos²theta - sin³theta ]/sin ²theta

>> [ (sin²theta)/(cos³theta) { cos theta - sin theta cos²theta + sin theta) ]/sin² theta

>> [ (sin²theta)/(cos³theta) { cos theta - sin theta( cos²theta - 1)} ]/sin² theta

>> [ (sin²theta)/(cos³theta) { cos theta + sin³theta]/sin²theta

>> 1/cos²theta + tan³theta

>> sec² theta + tan³ theta

[ The question might be wrong ]