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प्रश्न
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
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उत्तर
LHS= `(sectheta- tan theta)/(sec theta + tan theta)`
= `(1/cos theta-sin theta/cos theta)/(1/cos theta+ sin theta/cos theta)`
=`((1-sin theta)/cos theta)/((1+ sin theta)/cos theta)`
=`(1-sin theta)/(1+ sin theta)`
=`((1-sin theta) (1+ sin theta))/( (1+ sin theta )(1+ sin theta)) {"Dividing the numerator and
denominator by"(1 + cos theta)}`
=`((1-sin^2 theta))/((1+ sin theta)^2)`
=`cos^2 theta/(1+ sin theta)^2`
= RHS
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities
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`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
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`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
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`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
Prove the following identity :
`cosecA + cotA = 1/(cosecA - cotA)`
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Find the value of `θ(0^circ < θ < 90^circ)` if :
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Prove the following identities.
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Prove the following identities.
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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 sec θ = `41/40`, then find values of sin θ, cot θ, cosec θ
