Advertisements
Advertisements
प्रश्न
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
Advertisements
उत्तर
LHS =`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))`
=`((sinA - sin B )( sinA + sinB )+ ( cos A - cosB )( cosA - cosB))/((cos A+ cos B )( sin A+ sinB))`
=` (sin^2 A - sin^2 B + cos^2 A - cos^2 B)/( (cos A + cos B )( sinA + sinB))`
=` 0/((cos A + cos B )( sin A + sinB ))`
=0
=RHS
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Prove the following trigonometric identities.
`(1 + cos A)/sin^2 A = 1/(1 - cos A)`
Prove the following trigonometric identities.
`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`
Prove the following trigonometric identities.
`(1 - tan^2 A)/(cot^2 A -1) = tan^2 A`
Prove the following trigonometric identities.
(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
Prove the following trigonometric identities.
`(tan A + tan B)/(cot A + cot B) = tan A tan B`
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
`(cos theta cosec theta - sin theta sec theta )/(costheta + sin theta) = cosec theta - sec theta`
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
Prove the following identity :
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`
Prove the following identity :
`(1 + tan^2θ)sinθcosθ = tanθ`
Prove the following identity :
`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
Without using trigonometric identity , show that :
`tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ = 1/sqrt(3)`
Prove that :(sinθ+cosecθ)2+(cosθ+ secθ)2 = 7 + tan2 θ+cot2 θ.
Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.
If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to
Prove that cot2θ × sec2θ = cot2θ + 1
If 1 + sin2α = 3 sinα cosα, then values of cot α are ______.
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
If tan θ = `x/y`, then cos θ is equal to ______.
