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Question
Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.
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Solution
L.H.S. = (sin θ + cos θ)(tan θ + cot θ)
= `(sin theta + cos theta)(sin theta/cos theta + costheta/sin theta)`
= `(sin theta + cos theta)((sin^2 theta + cos^2 theta)/(costhetasin theta))`
= `(sintheta+costheta)xx1/(sinthetacostheta)` ...[∵ sin2θ + cos2θ = 1]
= `(sin theta + cos theta)/(cos theta sin theta)`
= `sin theta/(cos thetasin theta) + cos theta/(cos theta sin theta)`
= `1/cos theta + 1/sin theta`
= `sec theta + cosec theta`
= R.H.S
Hence proved.
RELATED QUESTIONS
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cosec θ – cot θ)^2 = (1-cos theta)/(1 + cos theta)`
Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
Prove the following trigonometric identities.
`sin theta/(1 - cos theta) = cosec theta + cot theta`
Prove the following trigonometric identities.
sin2 A cot2 A + cos2 A tan2 A = 1
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
`cot^2 theta - 1/(sin^2 theta ) = -1`a
`tan theta /((1 - cot theta )) + cot theta /((1 - tan theta)) = (1+ sec theta cosec theta)`
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
If`( 2 sin theta + 3 cos theta) =2 , " prove that " (3 sin theta - 2 cos theta) = +- 3.`
Write the value of `(1 + cot^2 theta ) sin^2 theta`.
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
Prove the following identity :
`sqrt((1 + cosA)/(1 - cosA)) = cosecA + cotA`
Prove the following identity :
`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
Prove that `"cot A"/(1 - cot"A") + "tan A"/(1 - tan "A")` = – 1
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
