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प्रश्न
Prove the following trigonometric identities.
`(1 + sin theta)/cos theta + cos theta/(1 + sin theta) = 2 sec theta`
उत्तर १
We have to prove `(1 + sin theta)/cos theta + cos theta/1+ sin theta - 2 sec theta`
We know that, `sin^2 theta + cos^2 theta = 1`
Multiplying the denominator and numerator of the second term by `(1 - sin theta)` we have
`(1 + sin theta)/cos theta + cos theta/(1 + sin theta) = (1 = sin theta)/cos theta = (cos theta(1 - sin theta))/((1 + sin theta)(1 - sin theta))`
`= (1 + sin theta)/cos theta = (cos theta (1 - sin theta))/(1-sin theta)`
`= (1 + sin theta)/cos theta + (cos theta(1 - sin theta))/cos^2 theta`
`= (1 + sin theta)/cos theta + (1 - sin theta)/cos theta`
`= (1 + sin theta + 1 -sin theta)/cos theta`
`= 2/cos theta`
`= 2 sec theta`
उत्तर २
LHS = `(1 + sin θ)/cos θ + cos θ/(1 + sin θ)`
= `(( 1 + sin θ)^2 + cos^2 θ)/(cos θ( 1 + sin θ))`
= `( 1 + sin^2 θ + 2 sin θ + cos^2 θ)/(cos θ( 1 + sin θ ))`
= `(1 + (sin^2θ + cos^2 θ) + 2 sin θ)/(cos θ(1 + sin θ))`
= `(1 + 1 + 2sin θ)/(cos θ(1 + sin θ))`
= `(2(1 + sin θ))/(cos θ(1 + sin θ))`
= 2 sec θ
Hence proved.
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Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= R.H.S
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Solution :
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= `cosθ/sinθ + sinθ/cosθ`
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= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
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