May 2026

If cos α + cos β = 1/3 and sin α + sin β = 1/4, Prove that cos((α − β)/2) = ± 5/24 If \[ \cos\alpha+\cos\beta=\frac{1}{3} \] and \[ \sin\alpha+\sin\beta=\frac{1}{4}, \] prove that \[ \cos\frac{\alpha-\beta}{2} = \pm\frac{5}{24} \] Question If \[ \cos\alpha+\cos\beta=\frac{1}{3} \] and \[ \sin\alpha+\sin\beta=\frac{1}{4}, \] prove that \[ \cos\frac{\alpha-\beta}{2} = \pm\frac{5}{24}. \] […]

If sec(x + α) + sec(x − α) = 2 sec x, Prove that cos x = ± √2 cos(α/2) If \[ \sec(x+\alpha)+\sec(x-\alpha)=2\sec x, \] prove that \[ \cos x=\pm \sqrt{2}\cos\frac{\alpha}{2} \] Question If \[ \sec(x+\alpha)+\sec(x-\alpha)=2\sec x, \] prove that \[ \cos x=\pm \sqrt{2}\cos\frac{\alpha}{2}. \] Solution Given, \[ \sec(x+\alpha)+\sec(x-\alpha)=2\sec x \] Converting secant into cosine,

If cos x = (cos α + cos β)/(1 + cos α cos β), Prove that tan(x/2) = ± tan(α/2) tan(β/2) If \[ \cos x= \frac{\cos\alpha+\cos\beta} {1+\cos\alpha\cos\beta}, \] prove that \[ \tan\frac{x}{2} = \pm \tan\frac{\alpha}{2} \tan\frac{\beta}{2} \] Question If \[ \cos x= \frac{\cos\alpha+\cos\beta} {1+\cos\alpha\cos\beta}, \] prove that \[ \tan\frac{x}{2} = \pm \tan\frac{\alpha}{2} \tan\frac{\beta}{2}. \] Solution

If 2 tan(α/2) = tan(β/2), Prove that cos α = (3 + 5 cos β) / (5 + 3 cos β) If \[ 2\tan\frac{\alpha}{2}=\tan\frac{\beta}{2}, \] prove that \[ \cos\alpha= \frac{3+5\cos\beta}{5+3\cos\beta} \] Question If \[ 2\tan\frac{\alpha}{2} = \tan\frac{\beta}{2}, \] prove that \[ \cos\alpha = \frac{3+5\cos\beta}{5+3\cos\beta}. \] Solution Given, \[ 2\tan\frac{\alpha}{2} = \tan\frac{\beta}{2} \] Therefore, \[ \tan\frac{\alpha}{2}

If sin α + sin β = a and cos α + cos β = b, Prove that cos(α − β) = (a² + b² − 2)/2 If sin α + sin β = a and cos α + cos β = b, prove that \[ \cos(\alpha-\beta)=\frac{a^2+b^2-2}{2} \] Question If \[ \sin\alpha+\sin\beta=a \] and \[

If sin α + sin β = a and cos α + cos β = b, Prove that sin(α + β) = 2ab/(a² + b²) If sin α + sin β = a and cos α + cos β = b, prove that \[ \sin(\alpha+\beta)=\frac{2ab}{a^2+b^2} \] Question If \[ \sin\alpha+\sin\beta=a \] and \[ \cos\alpha+\cos\beta=b, \]

If 2 tan α = 3 tan β, Prove that tan(α − β) = sin 2β / (5 − cos 2β) If 2 tan α = 3 tan β, prove that \[ \tan(\alpha-\beta)=\frac{\sin2\beta}{5-\cos2\beta} \] Question If \[ 2\tan\alpha = 3\tan\beta, \] prove that \[ \tan(\alpha-\beta) = \frac{\sin2\beta}{5-\cos2\beta}. \] Solution Given, \[ 2\tan\alpha = 3\tan\beta \]

Prove that cos(π/65) cos(2π/65) cos(4π/65) cos(8π/65) cos(16π/65) cos(32π/65) = 1/64 Prove that: \[ \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64} \] Question Prove that \[ \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64} \] Solution Using the identity \[ 2\sin\theta\cos\theta=\sin2\theta \] Start with \[ \cos\frac{\pi}{65} \cos\frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} \] Multiply and divide

Prove that cos(π/5) cos(2π/5) cos(4π/5) cos(8π/5) = -1/16 Prove that: cos(π/5) cos(2π/5) cos(4π/5) cos(8π/5) = -1/16 Question Prove that \[ \cos\frac{\pi}{5} \cos\frac{2\pi}{5} \cos\frac{4\pi}{5} \cos\frac{8\pi}{5} = -\frac{1}{16} \] Solution Using the identity \[ 2\sin\theta\cos\theta=\sin2\theta \] Start with \[ \cos\frac{\pi}{5} \cos\frac{2\pi}{5} \cos\frac{4\pi}{5} \cos\frac{8\pi}{5} \] Multiply and divide by \[ \sin\frac{\pi}{5} \] \[ = \frac{ \sin\frac{\pi}{5} \cos\frac{\pi}{5} \cos\frac{2\pi}{5}

Prove that cos(2π/15) cos(4π/15) cos(8π/15) cos(16π/15) = 1/16 Prove that: cos(2π/15) cos(4π/15) cos(8π/15) cos(16π/15) = 1/16 Question Prove that \[ \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \cos\frac{16\pi}{15} = \frac{1}{16} \] Solution Using the identity \[ 2\sin\theta\cos\theta=\sin2\theta \] Start with \[ \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \cos\frac{16\pi}{15} \] Multiply and divide by \[ \sin\frac{2\pi}{15} \] \[ = \frac{ \sin\frac{2\pi}{15} \cos\frac{2\pi}{15} \cos\frac{4\pi}{15}