Advertisements
Advertisements
प्रश्न
Prove that `(tan^2 theta - 1)/(tan^2 theta + 1)` = 1 – 2 cos2θ
Advertisements
उत्तर
L.H.S = `(tan^2 theta - 1)/(tan^2 theta + 1)`
= `(tan^2 theta - 1)/(sec^2 theta)`
= `(sin^2 theta)/(cos^2 theta) - 1 ÷ 1/(cos^2 theta)`
= `(sin^2 theta - cos^2 theta)/(cos^2 theta) xx (cos^2 theta)/1`
= sin2θ − cos2θ
= 1 − cos2θ − cos2θ
= 1 – cos2θ
L.H.S = R.H.S
Hence it is proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
tan2θ cos2θ = 1 − cos2θ
Prove the following identities:
`cot^2A((secA - 1)/(1 + sinA)) + sec^2A((sinA - 1)/(1 + secA)) = 0`
Prove that
`cot^2A-cot^2B=(cos^2A-cos^2B)/(sin^2Asin^2B)=cosec^2A-cosec^2B`
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
Prove that `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ = 3`.
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
Prove that `1/("cosec" theta - cot theta)` = cosec θ + cot θ
If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.
tan θ × `sqrt(1 - sin^2 θ)` 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`
