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प्रश्न
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
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उत्तर
LHS = `( tan A + sec A - 1)/(tan A - sec A + 1)`
= `(( tan A + sec A) - (sec^2 A - tan^2 A))/((tan A - sec A) + 1)`
= `(( tan A + sec A)( 1 - sec A + tan A))/(tan A - sec A + 1)`
= tan A + sec A
= `sin A/cos A + 1/cos A = (1 + sin A)/cos A`
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(tan theta)/(1-cot theta) + (cot theta)/(1-tan theta) = 1+secthetacosectheta`
[Hint: Write the expression in terms of sinθ and cosθ]
Prove the following identities:
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
Write the value of `(1 + tan^2 theta ) cos^2 theta`.
If 5 `tan theta = 4,"write the value of" ((cos theta - sintheta))/(( cos theta + sin theta))`
If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
The value of sin2θ + `1/(1 + tan^2 theta)` is equal to
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
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`
