Question

${\cos }^2{1}^o+{\cos }^2{2}^o+{\cos }^2{3}^o............+{\cos }^2{90}^o$=

• A. $0$
• B. $1$
• C. $45$
• D. $\frac{89}{2}$

Answer 1

The correct option is C $45$

We have,

${\cos }^2{1}^o+{\cos }^2{2}^o+{\cos }^2{3}^o+.......+{\cos }^2{90}^o$

Now, we can written that,

${\cos }^2{1}^o+{\cos }^2{2}^o+{\cos }^2{3}^o+.......+{\cos }^2{87}^o+{\cos }^2{88}^o+{\cos }^2{89}^o+{\cos }^2{90}^o$

$\Rightarrow {\cos }^2{1}^o+{\cos }^2{2}^o+{\cos }^2{3}^o+.......+{\cos }^2({90}^o-3^o)+{\cos }^2({90}^o-2^o)+{\cos }^2({90}^o-1^o)+{\cos }^2{90}^o$

$\Rightarrow {\cos }^2{1}^o+{\cos }^2{2}^o+{\cos }^2{3}^o+.......+{\sin }^2{3}^o+{\sin }^2{2}^o+{\sin }^2{1}^o+{\cos }^2{90}^o\therefore \cos ({90}^o-\theta )=\sin \theta$

$\Rightarrow {\cos }^2{1}^o+{\sin }^2{1}^o+{\cos }^2{2}^o+{\sin }^2{2}^o+{\cos }^2{3}^o+{\sin }^2{3}^o+.......+{\cos }^2{90}^o$

$\Rightarrow 1+1+1+...45times$

$\Rightarrow 45$

Hence, this is the answer.