Question

(sin theta + cos theta whole square )is equals to

Answer 1

Answer:

$(sin\theta+cos\theta)^{2}$

=$1+sin2\theta$

Step-by-step explanation:

Given $(sin\theta+cos\theta)^{2}$

=$sin^{2}\theta+cos^{2}\theta+2sin\theta \times cos\theta$

/*By algebraic identity:

$\boxed {a^{2}+b^{2}+2ab=(a+b)^{2}}$*/

=$1+2sin\theta \times cos\theta$

/* By Trigonometric identity:

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

= $1+sin2\theta$

\* Since ,

$\boxed {2sin\theta\times cos\theta = sin2\theta }$*/

Therefore,

$(sin\theta+cos\theta)^{2}$

=$1+sin2\theta$

••♪

Answer 2

Answer:

(sin theta+ cos theta)^2

= sin ^2 theta + cos ^2 theta

+ 2 sin theta cos theta

= 1+ sin 2 theta.

one more question for u

$If \: tan \theta \: + \frac{1}{tan\theta} = 2\\ find: \\ (1)sin\theta + cos\theta \\ (2)sec\theta + tan\theta$