Given $\frac{b+c}{11} = \frac{c+a}{12}= \frac{a+b}{13} $ for a $ \Delta ABC$ with
usual notation. If $ \frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma} $ , then the ordered triad $( \alpha, \beta, \gamma)$ has a value :
- (3, 4, 5)
- (19, 7, 25)
- (7, 19, 25)
- (5, 12, 13)
The Correct Option is C
Solution and Explanation
$\Rightarrow \; a = 7 \lambda , b = 6 \lambda , c = 5\lambda$
(using cosine formula)
$\cos A =\frac{1}{5} , \cos B = \frac{19}{35} , \cos C = \frac{5}{7} $
$\alpha: \beta:\gamma \Rightarrow 7 : 19 : 25 $
Top Questions on Trigonometric Identities
- The value of \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \) is:
- JEE Main - 2025
- Mathematics
- Trigonometric Identities
- If \( \tan^{-1} (\sqrt{\cos \alpha}) - \cot^{-1 (\cos \alpha) = x \), then what is \( \sin \alpha \)?}
- MHT CET - 2025
- Mathematics
- Trigonometric Identities
- If $ \tan \theta + \cot \theta = 4 $, then find the value of $ \tan^3 \theta + \cot^3 \theta $.
- BITSAT - 2025
- Mathematics
- Trigonometric Identities
- Find the value of $ \sin 75^\circ \cos 15^\circ + \cos 75^\circ \sin 15^\circ $.
- BITSAT - 2025
- Mathematics
- Trigonometric Identities
- $\tan\left(\frac{2\pi}{7}\right)\tan\left(\frac{4\pi}{7}\right) + \tan\left(\frac{4\pi}{7}\right)\tan\left(\frac{\pi}{7}\right) + \tan\left(\frac{\pi}{7}\right)\tan\left(\frac{2\pi}{7}\right) =$
- AP EAPCET - 2025
- Mathematics
- Trigonometric Identities
Questions Asked in JEE Main exam
- Find the equivalent resistance between two ends of the following circuit:
- JEE Main - 2025
- Current electricity
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R):
Assertion (A): In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases.
Reason (R): Free expansion of an ideal gas is an irreversible and an adiabatic process.In the light of the above statements, choose the correct answer from the options given below:
- JEE Main - 2025
- Current electricity
- Find the output voltage in the given circuit.
- JEE Main - 2025
- Current electricity
Current passing through a wire as function of time is given as $I(t)=0.02 \mathrm{t}+0.01 \mathrm{~A}$. The charge that will flow through the wire from $t=1 \mathrm{~s}$ to $\mathrm{t}=2 \mathrm{~s}$ is:
- JEE Main - 2025
- Current electricity
- A wire of length $ 25 \, \text{m} $ and cross-sectional area $ 5 \, \text{mm}^2 $ having resistivity $ 2 \times 10^{-6} \, \Omega \cdot \text{m} $ is bent into a complete circle. The resistance between diametrically opposite points will be:
- JEE Main - 2025
- Current electricity
Concepts Used:
Trigonometric Identities
Various trigonometric identities are as follows:
Even and Odd Functions
Cosecant and Secant are even functions, all the others are odd.
- sin (-A) = – sinA,
- cos (-A) = cos A,
- cosec (-A) = -cosec A,
- cot (-A) = -cot A,
- tan (-A) = – tan A,
- sec (-A) = sec A.
Pythagorean Identities
- sin2θ + cos2θ = 1
- 1 + tan2θ = sec2θ
- 1 + cot2θ = cosec2θ
Periodic Functions
- T-Ratios of (2π + x)
sin (2π + x) = sin x,
cos (2π + x) = cos x,
tan (2π + x) = tan x,
cosec (2π + x) = cosec x,
sec (2π + x) = sec x,
cot (2π+x)=cotx. - T-Ratios of (π -x)
sin (π–x) = sin x,
cos (π–x) = - cos x,
tan (π–x) = - tan x,
cosec (π–x) = cosec x,
sec (π–x) = - sec x,
cot (π–x) = - cot x. - T-Ratios of (π+ x)
sin (π+x) = - sin x,
cos (π+x) = - cos x,
tan (π+x) = tan x,
cosec (π+x) = - cosec x,
sec (π+x) = - sec x,
cot (π+x) = cot x. - T-Ratios of (2π – x)
sin (2π–x) = - sin x,
cos (2n–x) = cos x,
tan (2π–x) = - tan x,
cosec (2π–x) = - cosec x,
sec (2π–x) = sec x,
cot (2π-x) = - cot x
Sum and Difference Identities
- T-Ratios of (x + y)
sin (x+y) = sinx.cosy + cosx.sin y
cos (x+y) = cosx.cosy – sinx.siny - T-Ratios of (x – y)
sin (x–y) = sinx.cosy – cos.x.sin y
cos (x-y) = cosx.cosy + sinx.siny
Product of T-ratios
- 2sinx cosy = sin(x+y) + sin(x–y)
- 2cosx siny = sin(x+y) – sin(x–y)
- 2 cosx cosy = cos(x+y) + cos(x–y)
- 2sinx.siny = cos(x–y) – cos(x+y)
T-Ratios of (2x)
sin2x = 2sin x cos x
cos 2x = cos2x – sin2x
= 2cos2x – 1
= 1 – 2sin2x
T-Ratios of (3x)
sin 3x = 3sinx – 4sin3x
cos 3x = 4cos3x – 3cosx