Advertisements
Advertisements
प्रश्न
Prove that:
(cosec A – sin A) (sec A – cos A) sec2 A = tan A
Advertisements
उत्तर
L.H.S. = (cosec A – sin A) (sec A – cos A) × sec2 A
= `(1/sinA - sinA)(1/cosA - cosA) xx sec^2A` ...`{∵ cosec theta = 1/sintheta, sectheta = 1/costheta, 1 - sin^2theta = cos^2theta, 1 - cos^2theta = sin^2theta}`
= `((1 - sin^2A)/sinA)((1 - cos^2A)/cosA) 1/(cos^2A)`
= `(cos^2A)/(sinA)*(sin^2A)/(cosA)*1/(cos^2A)`
= `sinA/cosA`
= tan A = R.H.S.
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`( i)sin^{2}A/cos^{2}A+\cos^{2}A/sin^{2}A=\frac{1}{sin^{2}Acos^{2}A)-2`
`(ii)\frac{cosA}{1tanA}+\sin^{2}A/(sinAcosA)=\sin A\text{}+\cos A`
`( iii)((1+sin\theta )^{2}+(1sin\theta)^{2})/cos^{2}\theta =2( \frac{1+sin^{2}\theta}{1-sin^{2}\theta } )`
Prove the following trigonometric identities.
`1/(1 + sin A) + 1/(1 - sin A) = 2sec^2 A`
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
Prove the following identities:
(sec A – cos A) (sec A + cos A) = sin2 A + tan2 A
Prove the following identities:
`1/(cosA + sinA) + 1/(cosA - sinA) = (2cosA)/(2cos^2A - 1)`
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
Prove that `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ)) = 1/(tan θ + cot θ)`
If 3 sin θ = 4 cos θ, then 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`
