Advertisements
Advertisements
प्रश्न
Prove that `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1) = (1 + cos "A")/sin "A"`
Advertisements
उत्तर
LHS = `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1)`
= `((cot "A" + "cosec A") - ("cosec"^2 "A" - cot^2 "A"))/(cot "A" - "cosec A" + 1)`
= `((cot "A" + "cosec A")("cosec A" + cot "A")("cosec A" - cot "A"))/(cot "A" - "cosec A" + 1)`
= `((cot "A" + "cosec A") [1 - "cosec A" - cot "A"])/(cot"A"-"cosec A"+1)`
= `((cot "A" + "cosec A") (1-"cosec A"+cot"A"))/(1-"cosec A"+cot"A")`
= cot A + cosec A
= `cos"A"/sin"A"+1/sin"A"=(cos"A"+1)/sin"A"`
= `(1+cos"A")/sin"A"`
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`1 + cot^2 theta/(1 + cosec theta) = cosec theta`
Prove the following identities:
cot2 A – cos2 A = cos2 A . cot2 A
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
Prove that:
(sin A + cos A) (sec A + cosec A) = 2 + sec A cosec A
Prove the following identity :
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
Evaluate:
`(tan 65^circ)/(cot 25^circ)`
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
tan (90 – θ) = ?
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
