Advertisements
Advertisements
प्रश्न
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
Advertisements
उत्तर
LHS = (sin θ + cos θ)(cosec θ – sec θ)
= `(sin θ + cos θ)(1/sin θ - 1/cos θ)`
= `(sin θ + cos θ)((cos θ - sin θ)/(sin θ * cos θ))`
= `(cos^2θ - sin^2θ)/(sinθ * cosθ)`
= `(1 - 2sin^2θ)/(sinθ*cosθ)`
= `1/(sinθ * cosθ) - (2 sin^2θ)/(sinθ * cosθ)`
= `cosec θ · sec θ - (2 sin^2 θ)/(sin θ * cos θ)`
= cosec θ · sec θ – 2 tan θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove that `sqrt(sec^2 theta + cosec^2 theta) = tan theta + cot theta`
Prove the following trigonometric identities.
sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B
`1/((1+ sin θ)) + 1/((1 - sin θ)) = 2 sec^2 θ`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
Write the value of `4 tan^2 theta - 4/ cos^2 theta`
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
Prove the following identity :
`(tanθ + sinθ)/(tanθ - sinθ) = (secθ + 1)/(secθ - 1)`
Prove that:
tan (55° + x) = cot (35° – x)
Prove that `(sec A)/(tan A + cot A) = sin A`.
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
