Question

the value of (costheta+sintheta)²+ (costheta-sintheta)²=_______​

Answer 1

Solution

We Have

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

Using this identities

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

We get

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

$\tt : \implies \: cos ^{2} \theta+ sin^{2} \theta + \cancel{ 2sin \theta cos \theta }+ cos ^{2} \theta+ sin^{2} \theta - \cancel{2sin \theta cos \theta}$

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

We know that

$\tt : \implies \: cos ^{2} \theta+ sin^{2} \theta = 1$

Using this identities we get

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

$\tt : \implies 1 + 1$

$\tt : \implies 2$

Hence answer is

$\tt : \implies 2$

Answer 2

$\huge\bf\underline\pink{☆Solution☆}$

$\huge\bf\underline{Formulas}$

$\sf\orange{ (a+b)² = a²+b²+2ab}$

$\sf\purple{(a-b)² = a²+b²-2ab}$

From Trignometric identities

$\sf\pink{cos²θ+sin²θ =1}$

_____________________________________________

(cosθ+sinθ)² + (cosθ- sinθ)²

= cos²θ+sin²θ+2cosθ+sinθ + cos²θ+sin²θ-2cosθ+sinθ

= (cos²θ+sin²θ)+(cos²θ+sin²θ) (Using trignometric identity)

= 1 + 1

= 2

Therefore cosθ+sinθ)² + (cosθ- sinθ)² = 2