Advertisements
Advertisements
प्रश्न
Prove that tan2Φ + cot2Φ + 2 = sec2Φ.cosec2Φ.
Advertisements
उत्तर
L.H.S. = tan2Φ + cot2Φ + 2
= tan2Φ + 1 + cot2Φ + 1
= sec2Φ + cosec2Φ
= `1/cos^2 Φ + 1/sin^2Φ`
= `(sin^2 Φ + cos^2 Φ)/(sin^2 Φ.cos^2Φ )`
= `1/(sin^2 Φ. cos^2 Φ )`
= cosec2Φ. sec2Φ
= R.H.S.
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
Prove that:
`tanA/(1 - cotA) + cotA/(1 - tanA) = secA "cosec" A + 1`
If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that : x2 + y2 + z2 = r2
`sqrt((1+cos theta)/(1-cos theta)) + sqrt((1-cos theta )/(1+ cos theta )) = 2 cosec theta`
\[\frac{x^2 - 1}{2x}\] is equal to
If m = a secA + b tanA and n = a tanA + b secA , prove that m2 - n2 = a2 - b2
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
tan (90 – θ) = ?
tan θ × `sqrt(1 - sin^2 θ)` is equal to:
