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प्रश्न
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
उत्तर
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
APPEARS IN
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