Advertisements
Advertisements
Question
Prove that:
`sin 18^circ/(cos 72^circ )+ sqrt(3)(tan 10^circ tan 30^circ tan 40^circ tan50^circ tan 80^circ) `
Solution
`"LHS" = sin 18^circ/(cos 72^circ )+ sqrt(3)(tan 10^circ tan 30^circ tan 40^circ tan50^circ tan 80^circ) `
`=(sin 18^circ)/sin(90^circ -72^circ) + sqrt(3) [cot(90^circ - 10^circ)xx1/sqrt(3)xxcot(90^circ - 40^circ )xxtan50^circ]`
`=(sin 18^circ)/(sin 18^circ) + sqrt(3) (cot 80^circxxcot 50^circ)xxtan 50^circxxtan 80^circ)/)`
`= 1 + (1/tan 80^circxx1/ tan 50^circxxtan 50^circxxtan 80^circ)`
= 2
= RHS
APPEARS IN
RELATED QUESTIONS
Without using trigonometric tables, evaluate :
`sec 11^circ/("cosec" 79^circ)`
Without using trigonometric tables, evaluate :
`tan 27^circ/cot 63^circ`
Without using trigonometric tables, prove that:
cosec 80° − sec 10° = 0
Without using trigonometric tables, prove that:
tan266° − cot224° = 0
Without using trigonometric tables, prove that:
(sin 65° + cos 25°)(sin 65° − cos 25°) = 0
Prove that:
`(2 "sin" 68^circ)/(cos 10^circ )- (2 cot 15^circ)/(5 tan 75^circ) = ((3 tan 45^circ t an 20^circ tan 40^circ tan 50^circ tan 70^circ)) /5= 1`
Prove that:
\[\frac{sin\theta \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]
Prove that:
\[cot\theta \tan\left( 90° - \theta \right) - \sec\left( 90° - \theta \right)cosec\theta + \sqrt{3}\tan12° \tan60° \tan78° = 2\]
Prove that:
cos1° cos2° cos3° ... cos180° = 0
If sec2A = cosec(A - 42°), where 2A is an acute angle, then find the value of A.
Prove that `(sin "A" - cos "A" + 1)/(sin "A" + cos "A" - 1) = 1/(sec "A" - tan "A")`
A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 30°. Find the speed of the boat in metres per minute [Use `sqrt3` = 1.732]
Given that `tan (θ_1 + θ_2) = (tan θ_1 + tan θ_2)/(1 - tan θ_1 tan θ_2)` Find (θ1 + θ2) when tan θ1 = `1/2 and tan θ_2 = 1/3`.
Solve : Sin2θ - 3sin θ + 2 = 0 .
Solve the following equation: `(cos^2θ - 3 cosθ + 2)/sin^2θ` = 1.
`(sin 20°50' + tan 67°40')/(cos 32°20' - sin 15°10')`
The maximum value of `1/(cosec alpha)` is ______.
Prove that:
`(cos^2 "A")/(cos "A" - sin "A") + (sin "A")/(1 - cot "A")` = sin A + cos A
