Advertisements
Advertisements
प्रश्न
If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn
Advertisements
उत्तर
Consider `(m^2 - n^2) = (tanA + sinA)^2 - (tanA - sinA)^2`
⇒ [tanA + sinA - (tanA - sinA)] [tanA + sinA + (tanA - sinA)]
⇒ [2sinA][2tanA] = 4sinAtanA
Now LHS = `(m^2 - n^2)^2 = (4sinAtanA)^2 = 16sin^2Atan^2A`
Also , RHS = 16mn = 16(tanA + sinA)(tanA - sinA)
⇒ RHS = 16mn = `16(tan^2A - sin^2A) = 16(sin^2A/cos^2A - sin^2A)`
⇒ `16sin^2A((1 - cos^2A)/cos^2A) = 16sin^2A(sin^2A/cos^2A) = 16sin^2Atan^2A`
Thus , `(m^2 - n^2)^2` = 16mn
APPEARS IN
संबंधित प्रश्न
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
Prove the following identities:
`((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA)) = 2cotA`
Prove that:
(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1
` tan^2 theta - 1/( cos^2 theta )=-1`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
Prove the following identity :
secA(1 - sinA)(secA + tanA) = 1
Find A if tan 2A = cot (A-24°).
If tan θ = 2, where θ is an acute angle, find the value of cos θ.
`(sin A)/(1 + cos A) + (1 + cos A)/(sin A)` = 2 cosec A
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
