Advertisements
Advertisements
प्रश्न
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Advertisements
उत्तर
LHS = sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ )
= sec θ. sec θ - tan θ. tan θ
= sec2θ - tan2θ
= 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities:
`(\text{i})\text{ }\frac{\sin \theta }{1-\cos \theta }=\text{cosec}\theta+\cot \theta `
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
`(sec^2 theta-1) cot ^2 theta=1`
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
Show that none of the following is an identity:
`sin^2 theta + sin theta = 2`
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/min.
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
