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प्रश्न
`(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))`
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उत्तर
We have , `(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) `
=`((sin theta + cos theta )^2 + (sin theta - cos theta)^2) /((sin theta - cos theta )(sin theta + cos theta))`
=`(sin^2 theta + cos ^2 theta + 2 sin theta cos theta + sin^2 theta + cos^2 theta -2 sin theta cos theta)/(sin^2 theta - cos ^2 theta)`
=`(1+1)/(sin^2 theta - cos^2 theta)`
=`2/(sin^2 theta - cos^2 theta)`
Again ,` 2/(sin^2 theta - cos^2 theta)`
=`2/(sin^2 theta -(1-sin^2 theta))`
=`2/(2 sin ^2 theta -1)`
APPEARS IN
संबंधित प्रश्न
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
Show that `sqrt((1-cos A)/(1 + cos A)) = sinA/(1 + cosA)`
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
`Prove the following trigonometric identities.
`(sec A - tan A)^2 = (1 - sin A)/(1 + sin A)`
Prove the following trigonometric identities.
`(1 + cos theta - sin^2 theta)/(sin theta (1 + cos theta)) = cot theta`
Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)
Show that one of the values of each member of this equality is sin α sin β sin γ
Prove the following identities:
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
`(1 + cot^2 theta ) sin^2 theta =1`
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
`sin^6 theta + cos^6 theta =1 -3 sin^2 theta cos^2 theta`
`(cos ec^theta + cot theta )/( cos ec theta - cot theta ) = (cosec theta + cot theta )^2 = 1+2 cot^2 theta + 2cosec theta cot theta`
If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
If tan θ × A = sin θ, then A = ?
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`
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
Prove the following identity:
(sin2θ – 1)(tan2θ + 1) + 1 = 0
