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प्रश्न
If tan A = n tan B and sin A = m sin B, prove that `cos^2A = (m^2 - 1)/(n^2 - 1)`
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उत्तर
Given that, tan A = n tan B and sin A = m sin B.
`=> n = tanA/tanB` and `m = sinA/sinB`
∴ `(m^2 - 1)/(n^2 - 1) = ((sinA/sinB)^2 - 1)/((tanA/tanB)^2 - 1)`
= `(sin^2A/sin^2B - 1/1)/(tan^2A/(tan^2B) - 1)`
= `((sin^2A - sin^2B).tan^2B)/(sin^2B.(tan^2A - tan^2B))`
= `((sin^2A - sin^2B)/tan^2B)/((tan^2A - tan^2B)/sin^2B)`
= `((sin^2A - sin^2B)sin^2B)/((sin^2A/cos^2A-sin^2B/cos^2B)cos^2Bsin^2B)`
= `(sin^2A - sin^2B)/(((sin^2A.cos^2B - sin^2B.cos^2A)/(cos^2A.cos^2B)) cos^2B)`
= `((sin^2A - sin^2B)cos^2A)/(sin^2A.cos^2B - sin^2B.cos^2A)`
= `((sin^2A - sin^2B)cos^2A)/(sin^2A(1 - sin^2B) - sin^2B (1 - sin^2A))`
= `((sin^2A - sin^2B)cos^2A)/(sin^2A - sin^2A.sin^2B - sin^2B + sin^2B.sin^2A)`
= `((sin^2A - sin^2B)cos^2A)/(sin^2A -sin^2B)`
= cos2 A
संबंधित प्रश्न
If secθ + tanθ = p, show that `(p^{2}-1)/(p^{2}+1)=\sin \theta`
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Prove the following identity :
`sec^2A + cosec^2A = sec^2Acosec^2A`
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If tan θ = `x/y`, then cos 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`
