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प्रश्न
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
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उत्तर
LHS =` 1+(cot^2 theta)/((1+ cosectheta))`
=`1+((cosec^2 theta-1))/((cosectheta++1)) (∵ cosec^2 theta - cot^2 theta =1)`
=`1+((cosectheta+1)(cosec theta-1))/((cosec theta +1))`
=`1+ (cosec theta -1)`
=` cosec theta`
=RHS
APPEARS IN
संबंधित प्रश्न
Prove that (cosec A – sin A)(sec A – cos A) sec2 A = tan A.
Prove the following trigonometric identities.
`cot theta - tan theta = (2 cos^2 theta - 1)/(sin theta cos theta)`
Prove the following trigonometric identities.
`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
Prove the following trigonometric identities.
`(cosec A)/(cosec A - 1) + (cosec A)/(cosec A = 1) = 2 sec^2 A`
Prove the following trigonometric identities.
`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`
Prove the following trigonometric identities.
`(cos A cosec A - sin A sec A)/(cos A + sin A) = cosec A - sec A`
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
(i)` (1-cos^2 theta )cosec^2theta = 1`
If x=a `cos^3 theta and y = b sin ^3 theta ," prove that " (x/a)^(2/3) + ( y/b)^(2/3) = 1.`
If `cot theta = 1/ sqrt(3) , "write the value of" ((1- cos^2 theta))/((2 -sin^2 theta))`
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is
If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
Prove the following identity :
`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
Prove that `"cosec" θ xx sqrt(1 - cos^2theta)` = 1
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
