Question

The value of ${\sin }^2{65}^o+{\sin }^2{25}^o+{\cos }^2{35}^o+{\cos }^2{55}^o$ is:

• A. $0$
• B. $1$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $2$

Give. $si{n}^2{65}^{\circ }+si{n}^2{25}^{\circ }+co{s}^2{35}^{\circ }+co{s}^2{55}^{\circ }...\left(1\right)$

we know that is the sum of two angle is complementary then

let $A+B={90}^{\circ }\Rightarrow si{n}^{2}A+si{n}^{2}B=co{s}^{2}A+co{s}^{2}B=1...\left(a\right)$

Now M eq.(1) by formula (a)

$(si{n}^2{65}^{\circ }+si{n}^2{25}^{\circ })+(co{s}^2{35}^{\circ }+co{s}^2{55}^{\circ })$

$\Rightarrow 1+1=2$ so right answer is $C\left(2\right)$.