Question

The value of $\frac{co{s}^2{26}^{\circ }+co{s}^2{64}^{\circ }}{si{n}^2{55}^{\circ }+si{n}^2{35}^{\circ }}$ is:

(find out, without using trignometric tables)

• A. 1
• B. -1
• C. 0
• D. not defined

Answer 1

The correct option is A

1

$\frac{co{s}^2{26}^{\circ }+co{s}^2{64}^{\circ }}{si{n}^2{55}^{\circ }+si{n}^2{35}^{\circ }}=\frac{co{s}^2{26}^{\circ }+si{n}^2({90}^{\circ }-{64}^{\circ })}{co{s}^2({90}^{\circ }-{55}^{\circ })+si{n}^2{35}^{\circ }}=\frac{co{s}^2{26}^{\circ }+si{n}^2{26}^{\circ }}{co{s}^2{35}^{\circ }+si{n}^2{35}^{\circ }}=\frac{1}{1}=1$

[As $si{n}^2\theta +co{s}^2\theta =1$]