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प्रश्न
Prove that : `1 - (cos^2 θ)/(1 + sin θ) = sin θ`.
उत्तर
LHS = `1 - (cos^2 θ)/(1 + sin θ)`
= `1 - (1 - sin^2 θ)/(1 + sin θ)`
= `1 - ((1 - sin θ)(1 + sin θ))/(1 + sin θ)`
= 1 - ( 1 - sin θ )
= 1 - 1 + sin θ
= sin θ
= RHS
Hence proved.
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संबंधित प्रश्न
Prove the following trigonometric identities:
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(sec2 θ − 1) (cosec2 θ − 1) = 1
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`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
`(sec^2 theta -1)(cosec^2 theta - 1)=1`
Show that none of the following is an identity:
(i) `cos^2theta + cos theta =1`
What is the value of \[\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\]
If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.
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Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
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∴ L.H.S. = R.H.S.
