Advertisements
Advertisements
प्रश्न
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
Advertisements
उत्तर
L.H.S = sec2θ + cosec2θ
= 1 + tan2θ + 1 + cot2θ .....[∵ sec2θ = 1 + tan2θ and cosec2θ = 1 + cot2θ]
= 2 + tan2θ + cot2θ .....(i)
R.H.S = sec2θ x cosec2θ
= (1 + tan2θ) x (1 + cot2θ) .....[∵ sec2θ = 1 + tan2θ and cosec2θ = 1 + cot2θ]
= 1 + cot2θ + tan2θ + tan2θ x cot2θ
= 1 + cot2θ + tan2θ + tan2θ x (1/tan2θ) ...... [∵ cot2θ = 1/tan2θ]
= 2 + tan2θ + cot2θ .......(ii)
From (i) and (ii)
sec2θ + cosec2θ = sec2θ x cosec2θ
APPEARS IN
संबंधित प्रश्न
Prove that `cosA/(1+sinA) + tan A = secA`
Prove the following trigonometric identities.
`"cosec" theta sqrt(1 - cos^2 theta) = 1`
Prove the following trigonometric identities.
`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
Prove the following trigonometric identities.
`sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`
If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = ± 3.
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a2 + b2 = m2 + n2
Prove the following identities:
sec2A + cosec2A = sec2A . cosec2A
Prove that:
`cosA/(1 + sinA) = secA - tanA`
`sin theta (1+ tan theta) + cos theta (1+ cot theta) = ( sectheta+ cosec theta)`
`(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) = 2/((sin^2 theta - cos ^2 theta)) = 2/((2 sin^2 theta -1))`
\[\frac{x^2 - 1}{2x}\] is equal to
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
Prove the following identity :
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
If x = asecθ + btanθ and y = atanθ + bsecθ , prove that `x^2 - y^2 = a^2 - b^2`
Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`
If tan θ × A = sin θ, then A = ?
Prove that `(sin θ + "cosec" θ)/(sin θ) = 2 + cot^2θ`.
If cosA + cos2A = 1, then sin2A + sin4A = 1.
Prove the following:
(sin α + cos α)(tan α + cot α) = sec α + cosec α
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`
