Question

tan²θ – sin²θ = tan²θ × sin²θ. For proof of this complete the activity given below.

Activity:

L.H.S = `square`

= `square (1 - (sin^2theta)/(tan^2theta))`

= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`

= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`

= `tan^2theta (1 - square)`

= `tan^2theta xx square`    .....[1 – cos²θ = sin²θ]

= R.H.S

Answer 1

L.H.S = tan²θ – sin²θ 

= `tan^2theta (1 - (sin^2theta)/(tan^2theta))`

= `tan^2theta (1 - (sin^2theta)/((sin^2theta)/(cos^2theta)))`

= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/sin^2theta)`

= `tan^2theta (1 - cos^2theta)`

= tan²θ × sin²θ     .....[1 – cos²θ = sin²θ]

= R.H.S