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प्रश्न
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
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उत्तर
LHS: `((sin^3θ)/(cos^3θ))/((1 + sin^2θ)/(cos^2θ)) + ((cos^3θ)/(sin^3θ))/((1 + cos^2θ)/(sin^2θ))`
= `((sin^3θ)/(cos^3θ))/(((cos^2θ + sin^2θ))/cos^2θ) + ((cos^3θ)/(sin^3θ))/(((sin^2θ + cos^2θ))/sin^2θ)`
= `sin^3θ/cosθ + cos^3θ/sinθ`
= `(sin^4θ + cos^4θ)/(cosθsinθ)`
= `((sin^2θ + cos^2θ)^2 - 2 sin^2θ cos^2θ)/(cosθ sinθ)`
= `(1 - 2 sin^2θ cos^2θ)/(cosθ sinθ)`
= `1/(cos θ sinθ) - (2 sin^2θcos^2θ)/(cosθ sinθ)`
= secθ cosec θ – 2 sinθ cosθ
= RHS
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
`sqrt((1+cos theta)/(1-cos theta)) + sqrt((1-cos theta )/(1+ cos theta )) = 2 cosec theta`
If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`
If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
Prove the following identity :
`(1 + sinA)/(1 - sinA) = (cosecA + 1)/(cosecA - 1)`
Without using trigonometric table , evaluate :
`(sin47^circ/cos43^circ)^2 - 4cos^2 45^circ + (cos43^circ/sin47^circ)^2`
tan θ cosec2 θ – tan θ is equal to
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
tan θ × `sqrt(1 - sin^2 θ)` is equal to:
