Advertisements
Advertisements
प्रश्न
Without using trigonometric tables, prove that:
sin248° + sin242° = 1
उत्तर
LHS = sin248° + sin242°
= sin2 (`90^circ - 42^circ`) + sin2 `42^circ`
= `cos^2 42^circ+ sin^2 42^circ`
= 1
= RHS
APPEARS IN
संबंधित प्रश्न
In the below given figure, a tower AB is 20 m high and BC, its shadow on the ground, is 20√3 m long. Find the sun’s altitude.
Evaluate without using trigonometric tables,
`sin^2 28^@ + sin^2 62^@ + tan^2 38^@ - cot^2 52^@ + 1/4 sec^2 30^@`
Without using trigonometric tables, evaluate
`sin^2 34^@ + sin^2 56^@ + 2tan 18^@ tan 72^@ - cot^2 30^@`
Without using trigonometric tables, evaluate :
`sin 16^circ/cos 74^circ`
Without using trigonometric tables, prove that:
cosec272° − tan218° = 1
Without using trigonometric tables, prove that:
tan266° − cot224° = 0
Without using trigonometric tables, prove that:
(sin72° + cos18°)(sin72° − cos18°) = 0
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 the following:
`1/(1+sin^2theta) + 1/(1+cos^2theta) + 1/(1+sec^2theta) + 1/(1+cosec^2theta) = 2`
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 .
From trigonometric table, write the values of sin 37°19'.
Solve the following equation: `(cos θ)/(1 - sin θ) + cos θ/(1 + sinθ) = 4`.
Solve the following equation: `(cos^2θ - 3 cosθ + 2)/sin^2θ` = 1.
Using trigonometric table evaluate the following:
tan 25°45' + cot 45°25'.
Using trigonometric table evaluate the following:
sin 64°42' + cos 42°20'
Using trigonometric table evaluate the following:
cos 64°42' - sin 42°20'
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