Advertisements
Advertisements
प्रश्न
Given `cos38^circ sec(90^circ - 2A) = 1` , Find the value of <A
Advertisements
उत्तर
`cos38^circ sec(90^circ - 2A) = 1`
⇒ `cos38^circcosec2A = 1`
⇒ `cos38^circ (1/(sin2A)) = 1`
⇒ `sin2A = cos(90 - 52^circ)`
⇒ `sin2A = sin52^circ`
⇒ `2A = 52^circ`
⇒ `A = 26^circ`
APPEARS IN
संबंधित प्रश्न
Prove that:
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
` tan^2 theta - 1/( cos^2 theta )=-1`
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
Prove the following identity :
`(1 + sinA)/(1 - sinA) = (cosecA + 1)/(cosecA - 1)`
Prove the following Identities :
`(cosecA)/(cotA+tanA)=cosA`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
Prove that `tan^3 θ/( 1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ) = sec θ. cosec θ - 2 sin θ cos θ.`
If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = `± sqrt(a^2 + b^2 - c^2)`
Prove that `(sintheta + tantheta)/cos theta` = tan θ(1 + sec θ)
Prove that cot2θ – tan2θ = cosec2θ – sec2θ
