Question

$\frac{sin{42}^{\circ }}{sec{48}^{\circ }}+\frac{cos{42}^{\circ }}{cosec{48}^{\circ }}-\frac{4}{3}si{n}^2{30}^{\circ }=$___

• A. $\frac{-2}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $1$

Answer 1

Answer: (B)

$\frac{sin{42}^{\circ }}{sec{48}^{\circ }}+\frac{cos{42}^{\circ }}{cosec{48}^{\circ }}-\frac{4}{3}si{n}^2{30}^{\circ }$

$=\frac{sin{42}^{\circ }}{sec({90}^{\circ }-{42}^{\circ })}+\frac{cos{42}^{\circ }}{cosec(90-{42}^{\circ })}-\left(\frac{4}{3}\right)\left(\frac{1}{2^2}\right)$

$=\frac{sin{42}^{\circ }}{cosec{42}^{\circ }}+\frac{cos{42}^{\circ }}{sec{42}^{\circ }}-\frac{4}{3}\left(\frac{1}{4}\right)(\because cosec({90}^{\circ }-\theta )=sec\theta ,sec({90}^{\circ }-\theta )=cosec\theta )$

$=si{n}^2{42}^{\circ }+co{s}^2{42}^{\circ }-\frac{1}{3}$

$=1-\frac{1}{3}(\because si{n}^2\theta +co{s}^2\theta =1)=\frac{2}{3}$

$\therefore ,\frac{sin{42}^{\circ }}{sec{48}^{\circ }}+\frac{cos{42}^{\circ }}{cosec{48}^{\circ }}-\frac{4}{3}si{n}^2{30}^{\circ }=\frac{2}{3}$.

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