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प्रश्न
Prove the following trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
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उत्तर
We need to prove `cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos `
Solving the L.H.S, we get
`cos A/(1 - tan A) + sin A/(1 - cot A)`
= `cos A/(1 - sin A/cos A) + sin A/(1 - cos A/sin A)`
`= cos A/((cos A - sin A)/cos A) + sin A/((sin A - cos A)/sin A)`
`= cos^2 A/(cos A - sin A) + (sin^2 A)/(sin A - cos A)`
`= (cos^2 A - sin^2 A)/(cos A - sin A)`
`= ((cos A + sin A)(cos A - sin A))/(cos A - sin A)` [using `a^2 - b^2 = (a + b)(a -b)`]
= cos A + sin A
= RHS
Hence proved.
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संबंधित प्रश्न
Prove that `sqrt((1 + cos theta)/(1 - cos theta)) + sqrt((1 - cos theta)/(1 + cos theta)) = 2 cosec theta`
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`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
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`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
`(sin A)/(1 + cos A) + (1 + cos A)/(sin A)` = 2 cosec A
cos 45° = ?
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ.
If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ.
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`
