Advertisements
Advertisements
प्रश्न
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Advertisements
उत्तर
`(sec theta - 1)/(sec theta + 1)`
`= (1/cos theta - 1)/(1/cos theta + 1)`
= `((1 - cos theta)/cos theta)/((1 + cos theta)/cos theta)`
`= (1 - cos theta)/(1 +cos theta)`
`= (1 - cos theta)/(1 + cos theta) xx (1 + cos theta)/(1+ cos theta)`
`= (1 - cos^2 theta)/(1 + cos theta)^2`
`= sin^2 theta/(1 + cos theta)^2`
`= [sin theta/(1 + cos theta)]^2`
=RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p2 – 1) = 2p
(1 + tan θ + sec θ) (1 + cot θ − cosec θ) = ______.
Prove the following trigonometric identities.
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
Prove the following trigonometric identities
sec4 A(1 − sin4 A) − 2 tan2 A = 1
Prove that:
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
If `cos theta = 7/25 , "write the value of" ( tan theta + cot theta).`
If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`.
If `cos B = 3/5 and (A + B) =- 90° ,`find the value of sin A.
Find the value of `(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)`
If sin θ − cos θ = 0 then the value of sin4θ + cos4θ
Without using trigonometric identity , show that :
`cos^2 25^circ + cos^2 65^circ = 1`
Prove that `(sin θ tan θ)/(1 - cos θ) = 1 + sec θ.`
If x sin3θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ , then show that x2 + y2 = 1.
If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
If tan θ × A = sin θ, then A = ?
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
Prove the following identity:
(sin2θ – 1)(tan2θ + 1) + 1 = 0
