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प्रश्न
Prove the following trigonometric identities.
`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`
Prove the following:
`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`
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उत्तर
`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`
Consider the LHS = `(1 + cos θ + sin θ)/(1 + cos θ - sin θ)`
`= ((1 + cos θ + sin θ)/(1 + cos θ - sin θ))((1 + cos θ + sin θ)/(1 + cos θ + sin θ))`
`= (1 + cos θ + sin θ)^2/((1 + cos θ)^2 sin^2 θ)`
`= (2 + 2(cos θ + sin θ + sin θ cos θ))/(2 cos^2 θ+ 2 cos θ)`
`= (2(1 + cos θ)(1 + sin θ))/(2 cos θ (1 + cos θ))`
`= (1 + sin θ)/cos θ`
= RHS
Hence proved
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`sqrt((1+sinA)/(1-sinA)) = secA + tanA`
Prove that:
(sin A + cos A) (sec A + cosec A) = 2 + sec A cosec A
`(1 + cot^2 theta ) sin^2 theta =1`
Show that none of the following is an identity:
`sin^2 theta + sin theta = 2`
Prove the following identity :
secA(1 - sinA)(secA + tanA) = 1
Without using trigonometric identity , show that :
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`
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.
Let α, β be such that π < α – β < 3π. If sin α + sin β = `-21/65` and cos α + cos β = `-27/65`, then the value of `cos (α - β)/2` is ______.
