Question

Prove that sin(A+B) sin(A-B) = cos²B - cos²A ​

Answer 1

Step-by-step explanation:

Sin(A+B) =sinAcosB +cosAsinB

Sin(A-B) =

Sin(A+B) *sin(A-B) = sinAcosB +cosAsinB * sinAcosB-cosAsinB

= sin^2Acos^2B - cos^2Asin^2B

=(1-cos^2A)cos^2B-cos^2A(1-cos^2B)

=Cos^2B - cos^2Acos^2B - cos^2A + cos^2Acos^2B

=Cos^2B - Cos^2A

Answer 2

${\bf{\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}}$

To Prove:

• $\sin (A+B) \times \sin (A-B) = \cos^2B - \cos^2A$

Proof:

$\text{We know that,} \\ \sin (A+B) = \sin A*\cos B + \cos A*\sin B,\,and \\ \sin (A-B) = \sin A*\cos B - \cos A*\sin B \\ Taking\,LHS\, \\ \implies \sin (A+B) \times \sin (A-B) \\ \implies \sin A*\cos B + \cos A*\sin B \times \sin A*\cos B - \cos A*\sin B \\ \implies \sin A*\cos B \times \sin A*\cos B - \sin A*\cos B \times \cos A*\sin B + \\ \cos A*\sin B \times \sin A*\cos B - \cos A*\sin B \times \cos A*\sin B \\ (\sin A*\cos B \times \cos A*\sin B \:gets\:cut.\:So,) \\ \implies (\sin A*\cos B)^2 - (\cos A*\sin B)^2 \\ \implies \sin^2A*\cos^2B - \cos^2A*\sin^2B \\ \implies (1-\cos^2A)*\cos^2B - \cos^2A*(1-\cos^2B) \\ \implies \cos^2B- \cos^2A*\cos^2B - \cos^2A + \cos^2A*\cos^2B \\ \implies \bold{ \cos^2B - \cos^2A } \\ =RHS \\ \\ Hence\:Proved\:!!!$

Formulae Used:

• $\sin (A+B) = \sin A*\cos B + \cos A*\sin B$
• $\sin (A-B) = \sin A*\cos B - \cos A*\sin B$
• $\sin^2\theta = 1 - \cos^2\theta$

Learn More:

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$