Question

(cos theta + sin theta)^2 + (cos theta - sin theta)^2 is equal to



Mathematics​

Answer 1

Answer:

equal to 2

Step-by-step explanation:

2(cos^2 thita+sin^2thita) = 2(1)=2

Answer 2

GIVEN :–

$\\ \implies \tt( \cos \theta + \sin\theta)^2 + ( \cos \theta - \sin \theta)^2$

TO FIND :–

• Value of given value = ?

SOLUTION :–

• Let the given value –

$\\ \implies \tt P = ( \cos \theta + \sin\theta)^2 + ( \cos \theta - \sin \theta)^2$

• Using identity –

$\\ \longrightarrow \tt (a+b)^2 = a^2 + {b}^{2} + 2ab$

• And –

$\\ \longrightarrow \tt (a - b)^2 = a^2 + {b}^{2} -2ab$

• So that –

$\\ \implies \tt P = [\cos^{2} \theta + \sin^2\theta +2 \cos \theta \sin \theta ]+[\cos^{2} \theta + \sin^2\theta - 2 \cos \theta \sin \theta ]$

$\\ \implies \tt P = \cos^{2} \theta + \sin^2\theta +2 \cos \theta \sin \theta+\cos^{2} \theta + \sin^2\theta - 2 \cos \theta \sin \theta$

$\\ \implies \tt P = (\cos^{2} \theta + \cos^{2} \theta)+(\sin^2\theta + \sin^2\theta)+(2 \cos \theta \sin \theta - 2 \cos \theta \sin \theta)$

$\\ \implies \tt P = 2\cos^{2} \theta +2\sin^2\theta+0$

$\\ \implies \tt P = 2\cos^{2} \theta +2\sin^2\theta$

• We know that –

$\\ \longrightarrow \tt \sin^2 \theta +\cos^2 \theta = 1$

• So that –

$\\ \implies \tt P = 2(1)$

$\\\large\implies{ \boxed{\tt P = 2}}$