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प्रश्न
`(sec theta -1 )/( sec theta +1) = ( sin ^2 theta)/( (1+ cos theta )^2)`
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उत्तर
LHS = `(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+costheta)`
=`((1-cos theta)(1+ cos theta))/((1+ cos theta)(1+ cos theta)) {"Dividing the numerator and
denominator by "(1+ cos theta)}`
=`(1- cos^2 theta)/((1+ cos theta )^2)`
=`(sin^2 theta)/((1+ cos theta) ^2)`
= RHS
APPEARS IN
संबंधित प्रश्न
If tanθ + sinθ = m and tanθ – sinθ = n, show that `m^2 – n^2 = 4\sqrt{mn}.`
Prove the following trigonometric identity.
`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
Prove the following trigonometric identities.
(1 + cot A − cosec A) (1 + tan A + sec A) = 2
Prove the following identities:
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
If m = ` ( cos theta - sin theta ) and n = ( cos theta + sin theta ) "then show that" sqrt(m/n) + sqrt(n/m) = 2/sqrt(1-tan^2 theta)`.
What is the value of \[\frac{\tan^2 \theta - \sec^2 \theta}{\cot^2 \theta - {cosec}^2 \theta}\]
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
prove that `1/(1 + cos(90^circ - A)) + 1/(1 - cos(90^circ - A)) = 2cosec^2(90^circ - A)`
Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ.
If sec θ = x + `1/(4"x"), x ≠ 0,` find (sec θ + tan θ)
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
Prove that : `tan"A"/(1 - cot"A") + cot"A"/(1 - tan"A") = sec"A".cosec"A" + 1`.
If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
If tan θ = `7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ......[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
The value of the expression [cosec(75° + θ) – sec(15° – θ) – tan(55° + θ) + cot(35° – θ)] is ______.
