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
Show that `sqrt((1+cosA)/(1-cosA)) = cosec A + cot A`
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`
If `cot theta = 1/ sqrt(3) , "write the value of" ((1- cos^2 theta))/((2 -sin^2 theta))`
Write the value of cosec2 (90° − θ) − tan2 θ.
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
Prove that `(sin 70°)/(cos 20°) + (cosec 20°)/(sec 70°) - 2 cos 70° xx cosec 20°` = 0.
Prove that cosec θ – cot θ = `sin theta/(1 + cos theta)`
