Question

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

Activity:

L.H.S. = `square`

= `square (1 - (sin^2θ)/(tan^2θ))`

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

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

= `tan^2θ (1 - square)`

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

= R.H.S.

Answer 1

L.H.S. = \[\boxed{\text{tan}^2θ - \text{sin}^2θ}\] 

= \[\boxed{\text{tan}^2θ} \left(1 - \frac{\text{sin}^2θ}{\text{tan}^2θ}\right)\]

= \[\tan^2\theta\left(1-\frac{\boxed{\sin^2\theta}}{\frac{\sin^2\theta}{\cos^2\theta}}\right)\]

= \[\tan^{2}\theta\left(1-\frac{\sin^{2}\theta}{1}\times\frac{\cos^{2}\theta}{\boxed{\sin^{2}\theta}}\right)\]

= \[\text{tan}^2θ \left(1 - \boxed{\text{cos}^2θ}\right)\]

= \[\text{tan}^2θ × \boxed{\text{sin}^2θ}\]   ...[1 – cos²θ = sin²θ]

= R.H.S.