Advertisements
Advertisements
प्रश्न
If m = a secA + b tanA and n = a tanA + b secA , prove that m2 - n2 = a2 - b2
Advertisements
उत्तर
Given , m = a secA + b tanA and n = a tanA + b secA
`m^2 - n^2 = (asecA + btanA)^2 - (atanA + bsecA)^2`
⇒ `a^2sec^2A + b^2tan^2A + 2ab secAtanA - (a^2tan^2A + b^2 sec^2A + 2ab secAtanA)`
⇒ `sec^2A(a^2 - b^2) + tan^2A(b^2 - a^2) = (a^2 - b^2) [sec^2A - tan^2A]`
⇒ `(a^2 - b^2) ["Since" sec^2A - tan^2A = 1]`
Hence , `m^2 - n^2 = a^2 - b^2`
APPEARS IN
संबंधित प्रश्न
Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
Prove the following trigonometric identity.
`cos^2 A + 1/(1 + cot^2 A) = 1`
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
Prove that `(sin θ + "cosec" θ)/(sin θ) = 2 + cot^2θ`.
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
(tan θ + 2)(2 tan θ + 1) = 5 tan θ + sec2θ.
The value of 2sinθ can be `a + 1/a`, where a is a positive number, and a ≠ 1.
sec θ when expressed in term of cot θ, is equal to ______.
