Advertisements
Advertisements
प्रश्न
Find the value of cos 2A, A lies in the first quadrant, when tan A `16/63`
Advertisements
उत्तर
cos 2A = `(1 - tan^2"A")/(1 + tan^2"A")`
= `(1 - (16/63)^2)/(1 + (16/63)^2`
= `((63)^2 - (16)^2)/((63)^2 + (16)^2`
= `((63 + 16) (63 - 16))/(3969 + 256)`
= `(79 xx 47)/4225`
= `3713/4225`
APPEARS IN
संबंधित प्रश्न
Find the values of `tan ((19pi)/3)`
Find the value of the trigonometric functions for the following:
sec θ = `13/5`, θ lies in the IV quadrant
Prove that `(cot(180^circ + theta) sin(90^circ - theta) cos(- theta))/(sin(270^circ + theta) tan(- theta) "cosec"(360^circ + theta))` = cos2θ cotθ
If sin x = `15/17` and cos y = `12/13, 0 < x < pi/2, 0 < y < pi/2`, find the value of cos(x − y)
Find cos(x − y), given that cos x = `- 4/5` with `pi < x < (3pi)/2` and sin y = `- 24/25` with `pi < y < (3pi)/2`
Prove that sin2(A + B) – sin2(A – B) = sin2A sin2B
Show that tan(45° − A) = `(1 - tan "A")/(1 + tan "A")`
Prove that cot(A + B) = `(cot "A" cot "B" - 1)/(cot "A" + cot "B")`
If cos θ = `1/2 ("a" + 1/"a")`, show that cos 3θ = `1/2 ("a"^3 + 1/"a"^3)`
Express the following as a sum or difference
sin 4x cos 2x
Express the following as a sum or difference
2 sin 10θ cos 2θ
Show that `(sin 8x cos x - sin 6x cos 3x)/(cos 2x cos x - sin 3x sin 4x)` = tan 2x
If A + B + C = 180°, prove that `tan "A"/2 tan "B"/2 + tan "B"/2 tan "C"/2 + tan "C"/2 tan "A"/2` = 1
If A + B + C = 180°, prove that sin A + sin B + sin C = `4 cos "A"/2 cos "B"/2 cos "C"/2`
If A + B + C = `pi/2`, prove the following sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
If ∆ABC is a right triangle and if ∠A = `pi/2` then prove that sin2 B + sin2 C = 1
Choose the correct alternative:
`1/(cos 80^circ) - sqrt(3)/(sin 80^circ)` =
Choose the correct alternative:
cos 1° + cos 2° + cos 3° + ... + cos 179° =
