Advertisements
Advertisements
प्रश्न
Prove the following identities:
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
Advertisements
उत्तर
L.H.S. = `(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA)`
= `((sinA + cosA)^2 + (sinA - cosA)^2)/((sinA - cosA)(sinA + cosA))`
= `(sin^2A + cos^2A + 2sinAcosA + sin^2A + cos^2A - 2sinA cosA)/(sin^2A - cos^2A)`
= `(2(sin^2A + cos^2A))/(sin^2A - cos^2A)`
= `2/(sin^2A - cos^2A)` ...[sin2A + cos2A = 1]
= `2/(sin^2A - cos^2A)`
= `2/(sin^2A - (1 - sin^2A))`
= `2/(2sin^2A - 1)` = R.H.S.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
tan2θ cos2θ = 1 − cos2θ
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
`{1/((sec^2 theta- cos^2 theta))+ 1/((cosec^2 theta - sin^2 theta))} ( sin^2 theta cos^2 theta) = (1- sin^2 theta cos ^2 theta)/(2+ sin^2 theta cos^2 theta)`
Write the value of `sin theta cos ( 90° - theta )+ cos theta sin ( 90° - theta )`.
Prove the following identity :
tanA+cotA=secAcosecA
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove the following identity :
`sqrt(cosec^2q - 1) = "cosq cosecq"`
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
(1 – cos2 A) is equal to ______.
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`
