Advertisements
Advertisements
Question
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
Advertisements
Solution
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.
RELATED QUESTIONS
if `cos theta = 5/13` where `theta` is an acute angle. Find the value of `sin theta`
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
Prove that:
cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
`cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta = 2 sec theta `
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
Prove the following identity :
`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
