Advertisements
Advertisements
प्रश्न
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b(a2 – 1) is equal to
विकल्प
2a
3a
0
2ab
Advertisements
उत्तर
2a
Explanation;
Hint:
b(a2 – 1) = (sec θ + cosec θ) [(sin θ + cos θ)2 – 1]
= `1/ cos theta + 1/sin theta` [sin2 θ + cos2 θ + 2 sin θ cos θ – 1]
= `[(sin theta + cos theta)/(sin theta cos theta)]` [1 + 2 sin θ cos θ – 1]
= `[(sin theta + cos theta)/(sin theta cos theta)] xx 2 sin theta cos theta`
= 2(sin θ + cos θ)
= 2a
APPEARS IN
संबंधित प्रश्न
Prove that:
2 sin2 A + cos4 A = 1 + sin4 A
Prove the following identities:
(1 + tan A + sec A) (1 + cot A – cosec A) = 2
`(1-cos^2theta) sec^2 theta = tan^2 theta`
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
`cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta = 2 sec theta `
What is the value of \[\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\]
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
If x sin3θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ , then show that x2 + y2 = 1.
Prove that 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0.
If cosA + cos2A = 1, then sin2A + sin4A = 1.
