Advertisements
Advertisements
प्रश्न
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
विकल्प
a2 – b2
b2 – a2
a2 + b2
b – a
Advertisements
उत्तर
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
संबंधित प्रश्न
Prove the following trigonometric identities.
(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
Prove that:
cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
If `cot theta = 1/ sqrt(3) , "write the value of" ((1- cos^2 theta))/((2 -sin^2 theta))`
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
Prove the following identity :
`sqrt((secq - 1)/(secq + 1)) + sqrt((secq + 1)/(secq - 1))` = 2 cosesq
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
If tan θ = `9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove that `"cot A"/(1 - cot"A") + "tan A"/(1 - tan "A")` = – 1
