May 2026

Prove that: cot x + cot (π/3 + x) – cot (π/3 – x) = 3 cot 3x

Prove that cotx + cot(π/3 + x) − cot(π/3 − x) = 3 cot3x Prove that: \[ \cot x+\cot\left(\frac{\pi}{3}+x\right) -\cot\left(\frac{\pi}{3}-x\right) = 3\cot 3x \] Solution Let \[ \cot x=t \] Using \[ \cot(A+B) = \frac{\cot A\cot B-1}{\cot A+\cot B} \] \[ \cot(A-B) = \frac{\cot A\cot B+1}{\cot B-\cot A} \] and \[ \cot\frac{\pi}{3}=\frac{1}{\sqrt{3}} \] we get […]

Prove that: cot x + cot (π/3 + x) – cot (π/3 – x) = 3 cot 3x Read More »

Prove that: tan x + tan (π/3 + x) – tan (π/3 – x) = 3 tan 3x

Prove that tanx + tan(π/3 + x) − tan(π/3 − x) = 3 tan3x Prove that: \[ \tan x+\tan\left(\frac{\pi}{3}+x\right) -\tan\left(\frac{\pi}{3}-x\right) = 3\tan 3x \] Solution Let \[ \tan x=t \] Using the formula \[ \tan(A\pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B} \] and \[ \tan\frac{\pi}{3}=\sqrt{3} \] we get \[ \tan\left(\frac{\pi}{3}+x\right)

Prove that: tan x + tan (π/3 + x) – tan (π/3 – x) = 3 tan 3x Read More »

Prove that: tan x tan (x + π/3) + tan x tan (x – π/3) + tan (x + π/3) tan (x – π/3) = – 3

Prove that tanx tan(x + π/3) + tanx tan(x − π/3) + tan(x + π/3) tan(x − π/3) = −3 Prove that: \[ \tan x \tan \left(x+\frac{\pi}{3}\right) +\tan x \tan \left(x-\frac{\pi}{3}\right) +\tan \left(x+\frac{\pi}{3}\right) \tan \left(x-\frac{\pi}{3}\right) =-3 \] Solution Let \[ \tan x = t \] Using the formula \[ \tan(A\pm B) = \frac{\tan A \pm

Prove that: tan x tan (x + π/3) + tan x tan (x – π/3) + tan (x + π/3) tan (x – π/3) = – 3 Read More »

Prove that: 4 (cos³ 10° + sin³ 20°) = 3 (cos 10° + sin 20°)

Prove that 4(cos³10° + sin³20°) = 3(cos10° + sin20°) Prove that: \[ 4(\cos^3 10^\circ + \sin^3 20^\circ) = 3(\cos 10^\circ + \sin 20^\circ) \] Solution We use the identity \[ 4a^3 – 3a = \cos 3\theta \quad \text{when } a=\cos\theta \] So, \[ 4\cos^3 10^\circ = 3\cos 10^\circ + \cos 30^\circ \] Also, using \[

Prove that: 4 (cos³ 10° + sin³ 20°) = 3 (cos 10° + sin 20°) Read More »

Class 11th Maths – RD Sharma Chapter 9 : Value of Trigonometric Functions at Multiples and Submultiples of an angle – Exercise 9.2 Solutions (Step-by-Step Guide)

Value of Trigonometric Functions at Multiples and Submultiples of an angle – Exercise 9.2 Solutions Prove that: sin 5x = 5 sin x – 20 sin³ x + 16 sin⁵ x  Watch Solution Prove that: 4 (cos³ 10° + sin³ 20°) = 3 (cos 10° + sin 20°) Watch Solution Prove that: cos³ x sin 3x

Class 11th Maths – RD Sharma Chapter 9 : Value of Trigonometric Functions at Multiples and Submultiples of an angle – Exercise 9.2 Solutions (Step-by-Step Guide) Read More »

If cos α + cos β = 0 = sin α + sin β, then prove that cos 2α + cos 2β = -2 cos(α + β).

If cos α + cos β = 0 and sin α + sin β = 0, Prove that cos 2α + cos 2β = −2 cos(α + β) If \[ \cos\alpha+\cos\beta=0 \] and \[ \sin\alpha+\sin\beta=0, \] prove that \[ \cos2\alpha+\cos2\beta=-2\cos(\alpha+\beta) \] Question If \[ \cos\alpha+\cos\beta=0 \] and \[ \sin\alpha+\sin\beta=0, \] prove that \[ \cos2\alpha+\cos2\beta=-2\cos(\alpha+\beta). \]

If cos α + cos β = 0 = sin α + sin β, then prove that cos 2α + cos 2β = -2 cos(α + β). Read More »

If a cos 2x + b sin 2x = c has α and β as its roots, then prove that (i) tan α + tan β = 2b / (a + c) (ii) tan α tan β = (c – a) / (c + a) (iii) tan(α + β) = b/a

If a cos 2x + b sin 2x = c Has Roots α and β, Prove tan α + tan β, tan α tan β and tan(α + β) If \[ a\cos2x+b\sin2x=c \] has roots \(\alpha\) and \(\beta\), prove that \[ (i)\quad \tan\alpha+\tan\beta=\frac{2b}{a+c} \] \[ (ii)\quad \tan\alpha\tan\beta=\frac{c-a}{c+a} \] \[ (iii)\quad \tan(\alpha+\beta)=\frac{b}{a} \] Question If \[

If a cos 2x + b sin 2x = c has α and β as its roots, then prove that (i) tan α + tan β = 2b / (a + c) (ii) tan α tan β = (c – a) / (c + a) (iii) tan(α + β) = b/a Read More »

If sin α = 4/5 and cos β = 5/13 , prove that cos((α – β)/2) = 8/√65

If sin α = 4/5 and cos β = 5/13, Prove that cos((α − β)/2) = 8/√65 If \[ \sin\alpha=\frac{4}{5} \] and \[ \cos\beta=\frac{5}{13}, \] prove that \[ \cos\frac{\alpha-\beta}{2} = \frac{8}{\sqrt{65}} \] Question If \[ \sin\alpha=\frac{4}{5} \] and \[ \cos\beta=\frac{5}{13}, \] prove that \[ \cos\frac{\alpha-\beta}{2} = \frac{8}{\sqrt{65}}. \] Solution Given, \[ \sin\alpha=\frac{4}{5} \] Using \[

If sin α = 4/5 and cos β = 5/13 , prove that cos((α – β)/2) = 8/√65 Read More »