Ravi Kant Kumar

Prove that: sin²B = sin²A + sin²(A−B) − 2 sinA cosB sin(A−B) Question Prove that: \[ \sin^2 B = \sin^2 A + \sin^2(A-B) – 2\sin A\cos B\sin(A-B) \] Proof R.H.S. \[ = \sin^2 A+\sin^2(A-B)-2\sin A\cos B\sin(A-B) \] \[ = \sin^2 A+(\sin A\cos B-\cos A\sin B)^2 \] \[ -2\sin A\cos B(\sin A\cos B-\cos A\sin B) \]

Prove that: sin(A−B)/sinA sinB + sin(B−C)/sinB sinC + sin(C−A)/sinC sinA = 0 Question Prove that: \[ \frac{\sin(A-B)}{\sin A\sin B} + \frac{\sin(B-C)}{\sin B\sin C} + \frac{\sin(C-A)}{\sin C\sin A} =0 \] Proof L.H.S. \[ = \frac{\sin(A-B)}{\sin A\sin B} + \frac{\sin(B-C)}{\sin B\sin C} + \frac{\sin(C-A)}{\sin C\sin A} \] Using the identity: \[ \sin(X-Y) = \sin X\cos Y-\cos X\sin

Prove that: sin(A−B)/cosAcosB + sin(B−C)/cosBcosC + sin(C−A)/cosCcosA = 0 Question Prove that: \[ \frac{\sin(A-B)}{\cos A\cos B} + \frac{\sin(B-C)}{\cos B\cos C} + \frac{\sin(C-A)}{\cos C\cos A} =0 \] Proof L.H.S. \[ = \frac{\sin(A-B)}{\cos A\cos B} + \frac{\sin(B-C)}{\cos B\cos C} + \frac{\sin(C-A)}{\cos C\cos A} \] Using the identity: \[ \sin(X-Y) = \sin X\cos Y-\cos X\sin Y \] we

Prove that: [sin(A+B) + sin(A−B)] / [cos(A+B) + cos(A−B)] = tan A Question Prove that: \[ \frac{\sin(A+B)+\sin(A-B)} {\cos(A+B)+\cos(A-B)} = \tan A \] Proof L.H.S. \[ = \frac{\sin(A+B)+\sin(A-B)} {\cos(A+B)+\cos(A-B)} \] Using the identities: \[ \sin C+\sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2} \] and \[ \cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \] we get: \[ = \frac{ 2\sin\frac{(A+B)+(A-B)}{2} \cos\frac{(A+B)-(A-B)}{2} }

Prove that: sin²(n+1)A − sin²nA = sin(2n+1)A sin A Question Prove that: \[ \sin^2(n+1)A-\sin^2 nA = \sin(2n+1)A\sin A \] Proof L.H.S. \[ = \sin^2(n+1)A-\sin^2 nA \] Using the identity: \[ \sin^2 C-\sin^2 D = (\sin C-\sin D)(\sin C+\sin D) \] \[ = [\sin(n+1)A-\sin nA] [\sin(n+1)A+\sin nA] \] Using the identities: \[ \sin C-\sin D =

Prove that: cos²(π/4) − sin²(π/12) = √3/4 Question Prove that: \[ \cos^2\left(\frac{\pi}{4}\right) – \sin^2\left(\frac{\pi}{12}\right) = \frac{\sqrt{3}}{4} \] Proof L.H.S. \[ = \cos^2\left(\frac{\pi}{4}\right) – \sin^2\left(\frac{\pi}{12}\right) \] \[ = \left(\cos\frac{\pi}{4}\right)^2 – \left(\sin\frac{\pi}{12}\right)^2 \] \[ = \left(\frac{1}{\sqrt{2}}\right)^2 – \left(\sin15^\circ\right)^2 \] \[ = \frac{1}{2} – \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^2 \] \[ = \frac{1}{2} – \frac{(\sqrt{3}-1)^2}{8} \] \[ = \frac{1}{2} – \frac{3+1-2\sqrt{3}}{8} \]

If tan A = m/(m−1) and tan B = 1/(2m−1), Prove that A − B = π/4 Question If \[ \tan A=\frac{m}{m-1} \] and \[ \tan B=\frac{1}{2m-1} \] prove that: \[ A-B=\frac{\pi}{4} \] Proof Using the identity: \[ \tan(A-B) = \frac{\tan A-\tan B} {1+\tan A\tan B} \] Substituting the given values: \[ \tan(A-B) = \frac{\frac{m}{m-1}-\frac{1}{2m-1}}

If tan A = 5/6 and tan B = 1/11, Prove that A + B = π/4 Question If \[ \tan A=\frac{5}{6} \] and \[ \tan B=\frac{1}{11} \] prove that: \[ A+B=\frac{\pi}{4} \] Proof Using the identity: \[ \tan(A+B) = \frac{\tan A+\tan B} {1-\tan A\tan B} \] Substituting the given values: \[ \tan(A+B) = \frac{\frac{5}{6}+\frac{1}{11}}

Prove that: (tan 69° + tan 66°)/(1 − tan 69° tan 66°) = −1 Question Prove that: \[ \frac{\tan 69^\circ+\tan 66^\circ} {1-\tan 69^\circ \tan 66^\circ} =-1 \] Proof Consider the left-hand side: \[ \frac{\tan 69^\circ+\tan 66^\circ} {1-\tan 69^\circ \tan 66^\circ} \] Using the identity: \[ \tan(A+B) = \frac{\tan A+\tan B} {1-\tan A\tan B} \] Let

Prove that: sin(3π/8 − θ) cos(π/8 + θ) + cos(3π/8 − θ) sin(π/8 + θ) = 1 Question Prove that: \[ \sin\left(\frac{3\pi}{8}-\theta\right)\cos\left(\frac{\pi}{8}+\theta\right) + \cos\left(\frac{3\pi}{8}-\theta\right)\sin\left(\frac{\pi}{8}+\theta\right) =1 \] Proof Consider the left-hand side: \[ \sin\left(\frac{3\pi}{8}-\theta\right)\cos\left(\frac{\pi}{8}+\theta\right) + \cos\left(\frac{3\pi}{8}-\theta\right)\sin\left(\frac{\pi}{8}+\theta\right) \] Using the identity: \[ \sin A\cos B+\cos A\sin B=\sin(A+B) \] Let \[ A=\frac{3\pi}{8}-\theta, \qquad B=\frac{\pi}{8}+\theta \] Then, \[ =