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प्रश्न
`(sin theta+1-cos theta)/(cos theta-1+sin theta) = (1+ sin theta)/(cos theta)`
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उत्तर
LHS= `(sin theta+1cos theta)/(cos theta-1+sin theta) `
=`((sin theta+1-cos theta)(sin theta+cos theta+1))/((cos theta -1 + sin theta)(sin theta + cos theta +1))`
=`((sin theta + 1 )^2 - cos^2 theta)/((sin theta + cos theta )^2 -1^2)`
=`(sin^2 theta +1+2 sin theta - cos^2 theta)/(sin^2 + cos^2 theta+2 sin theta cos theta -1)`
=`(sin^2 theta + sin^2 theta + cos^2 theta +2sin theta - cos^2 theta)/(2 sin theta cos theta)`
=`(2 sin ^2 theta + 2 sin theta)/(2 sin theta cos theta)`
=`(2 sin theta (1+ sin theta))/(2 sin theta cos theta)`
=`(1+sin theta)/cos theta`
= RHS
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
if `T_n = sin^n theta + cos^n theta`, prove that `(T_3 - T_5)/T_1 = (T_5 - T_7)/T_3`
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
Prove the following identities:
`secA/(secA + 1) + secA/(secA - 1) = 2cosec^2A`
`sin^6 theta + cos^6 theta =1 -3 sin^2 theta cos^2 theta`
If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =
Prove the following identity :
secA(1 - sinA)(secA + tanA) = 1
Prove that:
`(cot A - 1)/(2 - sec^2 A) = cot A/(1 + tan A)`
Prove that `(cos θ)/(1 - sin θ) = (1 + sin θ)/(cos θ)`.
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
Prove that `sqrt((1 + sin θ)/(1 - sin θ))` = sec θ + tan θ.
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ
Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
If tan θ = `9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.
