Advertisements
Advertisements
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`
Advertisements
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
APPEARS IN
संबंधित प्रश्न
Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.
Prove the following trigonometric identity:
`sqrt((1 + sin A)/(1 - sin A)) = sec A + tan A`
Prove the following trigonometric identities.
`(cos theta)/(cosec theta + 1) + (cos theta)/(cosec theta - 1) = 2 tan theta`
Prove the following identities:
sec2A + cosec2A = sec2A . cosec2A
Prove the following identities:
`cosecA + cotA = 1/(cosecA - cotA)`
Prove the following identities:
`1/(1 + cosA) + 1/(1 - cosA) = 2cosec^2A`
Prove the following identities:
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
If sin A + cos A = p and sec A + cosec A = q, then prove that : q(p2 – 1) = 2p.
Prove that:
(sin A + cos A) (sec A + cosec A) = 2 + sec A cosec A
If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`
Simplify : 2 sin30 + 3 tan45.
sec4 A − sec2 A is equal to
If cos (\[\alpha + \beta\]= 0 , then sin \[\left( \alpha - \beta \right)\] can be reduced to
Prove the following identity :
`((1 + tan^2A)cotA)/(cosec^2A) = tanA`
Without using trigonometric table , evaluate :
`(sin47^circ/cos43^circ)^2 - 4cos^2 45^circ + (cos43^circ/sin47^circ)^2`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
Prove that `sqrt((1 + sin A)/(1 - sin A))` = sec A + tan A.
Prove that:
`sqrt((sectheta - 1)/(sec theta + 1)) + sqrt((sectheta + 1)/(sectheta - 1)) = 2cosectheta`
Prove that: 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1 = 0.
Prove that `sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A - 1) = 1`.
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
Prove the following identities.
cot θ + tan θ = sec θ cosec θ
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
Prove that
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A
If sin θ + cos θ = `sqrt(3)`, then show that tan θ + cot θ = 1
The value of 2sinθ can be `a + 1/a`, where a is a positive number, and a ≠ 1.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
