Math, asked by Mathematics3724, 1 year ago

Sin(90°-theta)cos(90°-theta)=tan theta/1+tan^2 theta

Answers

Answered by Anonymous
2

Step-by-step explanation:

sin(90'-theta)*cos(90'-theta)

=(sin90'*cos theta-cos90'*sin theta)(cos90'*cos theta + sin90'*sin theta)

= (1*cos theta-0)(0+1*sin theta)

= cos theta * sin theta

= 2 sin theta*cos theta/2

= sin (2theta)/2

= 1/2(2tan theta/1+tan² theta)

= tan theta/1+tan² theta

= tan theta/sec² theta

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