Question

(sin alpha cos beta + cos alpha sin beta ) square + (cos alpha cos beta - sin apha sin beta ) square= 1​

Answer 1

Question :-

Prove that :

$\bf{(sin\alpha cos\beta + cos\alpha sin\beta)^{2} + (cos\alpha cos\beta - sin\alpha sin \beta)^{2} = 1}$

To Find :-

To Proof that LHS = RHS.

We Know :-

• $\bf{sin(A + B) = sinAcosB + cosAsinB}$

• $\bf{cos(A + B) = cosAcosB - sinAsinB}$

• $\bf{sin^{2}\theta + cos^{2}\theta = 1}$

Concept :-

In the Equation,

LHS = $\bf{(sin\alpha cos\beta + cos\alpha sin\beta)^{2} + (cos\alpha cos\beta - sin\alpha sin \beta)^{2}}$

RHS = $\bf{1}$

So by solving the LHS , we can get the value of LHS.

Solution :-

$:\implies \bf{(sin\alpha cos\beta + cos\alpha sin\beta)^{2} + (cos\alpha cos\beta - sin\alpha sin \beta)^{2}}$

Now, by Using the formula of sin(A + B) and cos(A + B) , and Substituting it in the Equation , we get :-

$:\implies \bf{[sin(\alpha + \beta)]^{2} + [cos(\alpha + \beta)]^{2}}$

Taking $\bf{\alpha + \beta}$ as a single variable $\bf{\gamma}$ ,and putting it in the Equation , we get :-

$:\implies \bf{sin^{2}\gamma + cos^{2}\gamma}$

Using the identity ,and substituting it in the Equation , we get :-

$\bf{sin^{2}\theta + cos^{2}\theta = 1}$

$:\implies \bf{1}$

$\therefore \purple{\bf{LHS = 1}}$

Hence , LHS = 1.

Now , by putting LHS and RHS together , we get :-

$\bf{(sin\alpha cos\beta + cos\alpha sin\beta)^{2} + (cos\alpha cos\beta - sin\alpha sin \beta)^{2} = 1}$

Putting the value of LHS in thr Equation , we get :-

$\bf{1 = 1}$

Hence , it is proved that LHS = RHS.