If cos(A + B) sin(C − D) = cos(A − B) sin(C + D), prove that tan A tan B tan C + tan D = 0 If \[ \cos(A+B)\sin(C-D) = \cos(A-B)\sin(C+D) \] prove that \[ \tan A\tan B\tan C+\tan D=0 \] Solution Given: \[ \cos(A+B)\sin(C-D) = \cos(A-B)\sin(C+D) \] Use identities: \[ \cos(X+Y)=\cos X\cos Y-\sin […]

If y sin Φ = x sin(2θ + Φ), prove that (x + y) cot(θ + Φ) = (y − x) cot θ If \[ y\sin\Phi=x\sin(2\theta+\Phi) \] prove that \[ (x+y)\cot(\theta+\Phi)=(y-x)\cot\theta \] Solution Given: \[ y\sin\Phi=x\sin(2\theta+\Phi) \] Use identity: \[ \sin(A+B)=\sin A\cos B+\cos A\sin B \] \[ y\sin\Phi = x[\sin2\theta\cos\Phi+\cos2\theta\sin\Phi] \] Use identities: \[ \sin2\theta=2\sin\theta\cos\theta

If cos(α + β) sin(γ + δ) = cos(α − β) sin(γ − δ), prove that cot α cot β cot γ = cot δ If \[ \cos(\alpha+\beta)\sin(\gamma+\delta) = \cos(\alpha-\beta)\sin(\gamma-\delta) \] prove that \[ \cot\alpha\cot\beta\cot\gamma=\cot\delta \] Solution Given: \[ \cos(\alpha+\beta)\sin(\gamma+\delta) = \cos(\alpha-\beta)\sin(\gamma-\delta) \] Use identities: \[ \cos X\sin Y = \frac12[\sin(X+Y)+\sin(Y-X)] \] \[ \frac12[ \sin(\alpha+\beta+\gamma+\delta)

If cos(A − B)/cos(A + B) + cos(C + D)/cos(C − D) = 0, prove that tan A tan B tan C tan D = −1 If \[ \frac{\cos(A-B)}{\cos(A+B)} + \frac{\cos(C+D)}{\cos(C-D)} =0 \] prove that \[ \tan A\tan B\tan C\tan D=-1 \] Solution Given: \[ \frac{\cos(A-B)}{\cos(A+B)} + \frac{\cos(C+D)}{\cos(C-D)} =0 \] Transpose second term: \[ \frac{\cos(A-B)}{\cos(A+B)}

Prove that sin(B − C) cos(A − D) + sin(C − A) cos(B − D) + sin(A − B) cos(C − D) = 0 Prove that: \[ \sin(B-C)\cos(A-D) \] \[ +\sin(C-A)\cos(B-D) \] \[ +\sin(A-B)\cos(C-D)=0 \] Solution L.H.S. \[ = \sin(B-C)\cos(A-D) \] \[ +\sin(C-A)\cos(B-D) \] \[ +\sin(A-B)\cos(C-D) \] Use identity: \[ \sin X\cos Y = \frac12[\sin(X+Y)+\sin(X-Y)]

Prove that {cos(A+B+C) + cos(−A+B+C) + cos(A−B+C) + cos(A+B−C)}/{sin(A+B+C) + sin(−A+B+C) + sin(A−B+C) + sin(A+B−C)} = cot C Prove that: \[ \frac{ \cos(A+B+C)+\cos(-A+B+C)+\cos(A-B+C)+\cos(A+B-C) }{ \sin(A+B+C)+\sin(-A+B+C)+\sin(A-B+C)+\sin(A+B-C) } = \cot C \] Solution L.H.S. \[ = \frac{ \cos(A+B+C)+\cos(-A+B+C) }{ \sin(A+B+C)+\sin(-A+B+C) } \] \[ + \frac{ \cos(A-B+C)+\cos(A+B-C) }{ \sin(A-B+C)+\sin(A+B-C) } \] Use identities: \[ \cos X+\cos Y =

If sin 2A = λ sin 2B, prove that tan(A + B)/tan(A − B) = (λ + 1)/(λ − 1) If \[ \sin2A=\lambda\sin2B \] prove that \[ \frac{\tan(A+B)}{\tan(A-B)} = \frac{\lambda+1}{\lambda-1} \] Solution Given: \[ \sin2A=\lambda\sin2B \] Write \(\sin2A\) and \(\sin2B\) using identities: \[ 2\sin A\cos A = \lambda(2\sin B\cos B) \] \[ \sin A\cos A

If cosec A + sec A = cosec B + sec B, prove that tan A tan B = cot(A + B)/2 If \[ \csc A+\sec A=\csc B+\sec B \] prove that \[ \tan A\tan B=\cot\frac{A+B}{2} \] Solution Given: \[ \csc A+\sec A=\csc B+\sec B \] Write in terms of sine and cosine: \[ \frac1{\sin

If cos A + cos B = 1/2 and sin A + sin B = 1/4, prove that tan(A + B)/2 = 1/2 If \[ \cos A+\cos B=\frac12 \] and \[ \sin A+\sin B=\frac14 \] prove that \[ \tan\frac{A+B}{2}=\frac12 \] Solution Given: \[ \cos A+\cos B=\frac12 \] \[ \sin A+\sin B=\frac14 \] Use identities: \[

Prove that cos(A + B + C) + cos(A − B + C) + cos(A + B − C) + cos(−A + B + C) = 4 cos A cos B cos C Prove that: \[ \cos(A+B+C)+\cos(A-B+C) \] \[ +\cos(A+B-C)+\cos(-A+B+C) \] \[ =4\cos A\cos B\cos C \] Solution L.H.S. \[ = \cos(A+B+C)+\cos(A-B+C) \] \[ +\cos(A+B-C)+\cos(-A+B+C)