Advertisements
Advertisements
प्रश्न
If sin 3A = 1 and 0 < A < 90°, find `tan^2A - (1)/(cos^2 "A")`
Advertisements
उत्तर
sin 3A = 1
sin 3A = sin90°
3A = 90°
A = 30°
`tan^2A – (1)/(cos^2"A") = tan^2 30° – (1)/(cos^2 30°)`
= `(1/sqrt3)^2 – (1)/(sqrt3/2)^2`
= `(1)/(3) – (4)/(3)`
= `(–3)/(3)`
= – 1
संबंधित प्रश्न
Solve the following equation for A, if sec 2A = 2
Solve the following equation for A, if `sqrt3` cot 2 A = 1
Find the value of 'A', if cosec 3A = `(2)/sqrt(3)`
Find the value of 'A', if (1 - cosec A)(2 - sec A) = 0
If sin α + cosβ = 1 and α= 90°, find the value of 'β'.
If θ = 30°, verify that: 1 - sin 2θ = (sinθ - cosθ)2
Find the value of: `sqrt((1 - sin^2 60°)/(1 + sin^2 60°)` If 3 tan2θ - 1 = 0, find the value
a. cosθ
b. sinθ
Evaluate the following: `(sec34°)/("cosec"56°)`
Evaluate the following: `(sec32° cot26°)/(tan64° "cosec"58°)`
Evaluate the following: `(5sec68°)/("cosec"22°) + (3sin52° sec38°)/(cot51° cot39°)`
