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
Prove that `cosA/(1+sinA) + tan A = secA`
Prove the following trigonometric identities
sec4 A(1 − sin4 A) − 2 tan2 A = 1
Prove that:
`1/(cosA + sinA - 1) + 1/(cosA + sinA + 1) = cosecA + secA`
If m = a sec A + b tan A and n = a tan A + b sec A, then prove that : m2 – n2 = a2 – b2
Prove that:
Sin4θ - cos4θ = 1 - 2cos2θ
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Without using trigonometric identity , show that :
`cos^2 25^circ + cos^2 65^circ = 1`
Prove that `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ)) = 1/(tan θ + cot θ)`
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
