Advertisements
Advertisements
प्रश्न
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
Advertisements
उत्तर
LHS= `(1+tan^2theta)(1+cot^2 theta)`
=`sec^2 theta. cosec^2 theta (∵ sec^2 theta - tan^2 theta=1 and cosec^2 - cot^2 theta =1)`
=`1/(cos^2 theta. sin^theta)`
=` 1/((1-sin^2 theta ) sin^2 theta`
=`1/(sin^2theta-sin^4theta)`
==RHS
Hence, LHS = RHS
APPEARS IN
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A`
Prove that (cosec A – sin A)(sec A – cos A) sec2 A = tan A.
Without using trigonometric tables evaluate
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80^@)`
Prove the following trigonometric identities
`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
Prove the following identities:
sec2 A . cosec2 A = tan2 A + cot2 A + 2
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
Prove that:
2 sin2 A + cos4 A = 1 + sin4 A
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
Prove the following identities:
`(sinA - cosA + 1)/(sinA + cosA - 1) = cosA/(1 - sinA)`
`(tan A + tanB )/(cot A + cot B) = tan A tan B`
If m = ` ( cos theta - sin theta ) and n = ( cos theta + sin theta ) "then show that" sqrt(m/n) + sqrt(n/m) = 2/sqrt(1-tan^2 theta)`.
Simplify : 2 sin30 + 3 tan45.
Prove the following identity :
`sec^4A - sec^2A = sin^2A/cos^4A`
If sinA + cosA = m and secA + cosecA = n , prove that n(m2 - 1) = 2m
Prove that: `(sin θ - 2sin^3 θ)/(2 cos^3 θ - cos θ) = tan θ`.
Prove that 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0
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.
