Advertisements
Advertisements
Question
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
Options
a2 – b2
b2 – a2
a2 + b2
b – a
Advertisements
Solution
b2 – a2
Explanation;
(a cot θ + b cosec θ)2 = p2
(b cot θ + a cosec θ)2 = q2
p2 – q2 = a2 cot2θ + a2 cot2θ + 2ab cot θ cosec θ – (b2 cot2θ + a2 cosec2θ + 2ab cot θ cosec θ)
= (a2 – b2) cot2θ + (b2 – a2) cosec2θ
= (a2 – b2) (cosec2θ – 1) + (b2 – a2) (cosec2θ)
= (a2 – b2) cosec2θ – (a2 – b2) – (a2 – b2) cosec2θ
= b2 – a2
APPEARS IN
RELATED QUESTIONS
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
Prove the following trigonometric identities.
`tan theta + 1/tan theta` = sec θ.cosec θ
Prove the following trigonometric identities.
`(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`
If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =
Find the value of ( sin2 33° + sin2 57°).
Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.
If sin θ + sin2 θ = 1 show that: cos2 θ + cos4 θ = 1
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
Eliminate θ if x = r cosθ and y = r sinθ.
