Advertisements
Advertisements
प्रश्न
Prove the following trigonometric identities.
`(cos A cosec A - sin A sec A)/(cos A + sin A) = cosec A - sec A`
Advertisements
उत्तर
We have to prove `(cos A cosec A - sin A sec A)/(cos A + sin A) = cosec A - sec A`
So,
`(cos A cosec A - sin A sec A)/(cos A + sin A) = (cos A 1/sin A - sin A 1/cos A)/(cos A + sin A)`
`= ((cos^2 A - sin^2 A)/(sin A cos A))/(cos A + sin A)`
`= (cos^2 A - sin^2 A)/(sin A cos A(cos A + sin A))`
`= ((cos A - sin A)(cos A + sin A))/(sin A cos A(cos A + sin A))`
`= (cos A - sin A)/(sin A cos A)`
`= cos A/(sin A cos A) - sin A/(sin A cos A)``
`= 1/sin A - 1/cos A``
`= cosec A - sec A`
Hence proved.
APPEARS IN
संबंधित प्रश्न
If sinθ + sin2 θ = 1, prove that cos2 θ + cos4 θ = 1
Prove the following trigonometric identities:
`(1 - cos^2 A) cosec^2 A = 1`
Prove the following trigonometric identities.
`(tan^3 theta)/(1 + tan^2 theta) + (cot^3 theta)/(1 + cot^2 theta) = sec theta cosec theta - 2 sin theta cos theta`
Prove the following identities:
`((1 + tan^2A)cotA)/(cosec^2A) = tan A`
Prove the following identities:
`(1 - sinA)/(1 + sinA) = (secA - tanA)^2`
Prove the following identities:
(1 + tan A + sec A) (1 + cot A – cosec A) = 2
Prove that:
(cosec A – sin A) (sec A – cos A) sec2 A = tan A
`sin^2 theta + cos^4 theta = cos^2 theta + sin^4 theta`
`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
What is the value of (1 − cos2 θ) cosec2 θ?
The value of \[\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}\]
If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =
(sec A + tan A) (1 − sin A) = ______.
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove the following identities:
`(sec"A"-1)/(sec"A"+1)=(sin"A"/(1+cos"A"))^2`
Prove the following identity :
`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
If x = r sinA cosB , y = r sinA sinB and z = r cosA , prove that `x^2 + y^2 + z^2 = r^2`
If sin θ + cos θ = `sqrt(3)`, then show that tan θ + cot θ = 1
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`
