Advertisements
Advertisements
Question
if `sec A = 5/4` verify that `(3 sin A - 4 sin^3 A)/(4 cos^3 A - 3 cos A) = (3 tan A - tan^3 A)/(1- 3 tan^2 A)`
Solution
We have
`sec A = 5/4`
In ΔABC
`AC^2 = AB^2 + BC^2`
`=> (5)^2 = (4)^2 + (BC)^2`
`=> BC^2 = 25 - 16`
=> BC = 3
`sin A = 3/5, cos A = 4/5 and tan A = 3/4`
Now
`(3sin A - 4sin^3 A)/(4 cos^3 A - 3 cosA) = (3 tan A - tan^3 A)/(1 - 3tan^2 A)`
`=> (3 xx 3/5 - 4 xx (3/5)^2)/(4 xx (4/5)^3 - 3 xx (4/5)) = (3xx (3/4) - (3/4)^3)/(1 - 3xx(3/4)^2)`
`=> (9/5 - 108/125)/(256/25 - 12/5) = (9/4 - 27/64)/(1 - 27/16)`
`=> ((225 - 108)/125)/((256 - 300)/125) = ((144 - 27)/64)/((16 - 27)/16)`
`=> 117/(-44) = 117/(-11xx4)`
=> L.H.S = R.H.S
APPEARS IN
RELATED QUESTIONS
If ЁЭЬГ = 30° verify `cos 2 theta = (1 - tan^2 theta)/(1 + tan^2 theta)`
In a ΔABC , ∠B = 90° , AB= 24 cm and BC = 7 cm find (i) sin A (ii) cos A (iii) sin C (iv) cos C
Evaluate:
`4/(cot^2 30^0) +1/(sin^2 30^0) -2 cos^2 45^0 - sin^2 0^0`
If cosec θ = `sqrt5`, find the value of:
- 2 - sin2 θ - cos2 θ
- 2 + `1/sin^2"θ" – cos^2"θ"/sin^2"θ"`
In the given figure, triangle ABC is right-angled at B. D is the foot of the perpendicular from B to AC. Given that BC = 3 cm and AB = 4 cm.
find :
- tan ∠DBC
- sin ∠DBA
In triangle ABC, AB = AC = 15 cm and BC = 18 cm, find cos ∠ABC.
If 5 cos θ = 3, evaluate : `(co secθ – cot θ)/(co secθ + cot θ)`
In each of the following, one trigonometric ratio is given. Find the values of the other trigonometric.
cose C = `(15)/(11)`
In each of the following, one trigonometric ratio is given. Find the values of the other trigonometric.
tanB = `(8)/(15)`
Evaluate: `5/(cot^2 30^circ) + 1/(sin^2 60^circ) - cot^2 45^circ + 2 sin^2 90^circ`.
