Question

$\frac{co{s}^2{56}^{\circ }+co{s}^2{34}^{\circ }}{si{n}^2{56}^{\circ }+si{n}^2{34}^{\circ }}+3ta{n}^2{56}^{\circ }\times ta{n}^2{34}^{\circ }=$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4

$\frac{co{s}^2{56}^{\circ }+co{s}^2{34}^{\circ }}{si{n}^2{56}^{\circ }+si{n}^2{34}^{\circ }}+3ta{n}^2{56}^{\circ }\times ta{n}^2{34}^{\circ }$
We know that,
$cos\theta =sin({90}^{\circ }-\theta )sin\theta =cos({90}^{\circ }-\theta )tan\theta =cot({90}^{\circ }-\theta )$
By using these identities, we get,

$=\frac{si{n}^2({90}^{\circ }-{56}^{\circ })+co{s}^2{34}^{\circ }}{co{s}^2({90}^{\circ }-{56}^{\circ })+si{n}^2{34}^{\circ }}+3co{t}^2({90}^{\circ }-{56}^{\circ })\times ta{n}^2{34}^{\circ }=\frac{si{n}^2{34}^{\circ }+co{s}^2{34}^{\circ }}{co{s}^2{34}^{\circ }+si{n}^2{34}^{\circ }}+3co{t}^2{34}^{\circ }\times ta{n}^2{34}^{\circ }$
Applying the trigonometric identity$si{n}^2\theta +co{s}^2\theta =1$ and the relation $cot\theta =\frac{1}{tan\theta }$,

$=\frac{1}{1}+3=4$