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.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Prove the following identities:
(cos A + sin A)2 + (cos A – sin A)2 = 2
If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`
If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
Prove the following identity :
`[1/((sec^2θ - cos^2θ)) + 1/((cosec^2θ - sin^2θ))](sin^2θcos^2θ) = (1 - sin^2θcos^2θ)/(2 + sin^2θcos^2θ)`
Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2.
sec 60° = ?
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
