Question

The value of $tan{48}^{\circ }.tan{23}^{\circ }.tan{42}^{\circ }.tan{67}^{\circ }$
and
$cos{38}^{\circ }.cos{52}^{\circ }-sin{38}^{\circ }.sin{52}^{\circ }$ is equal to

• A. 1 and 0 respectively
• B. 1 and 1 respectively
• C. 1 and 2 respectively
• D. 2 and 1 respectively

Answer 1

The correct option is A 1 and 0 respectively

i) $tan{48}^{\circ }.tan{23}^{\circ }.tan{42}^{\circ }.tan{67}^{\circ }$

We know that, $cot({90}^{\circ }-\theta )=tan\theta$

$\therefore tan{48}^{\circ }.tan{23}^{\circ }.tan{42}^{\circ }.tan{67}^{\circ }$
= $tan{48}^{\circ }.tan{23}^{\circ }.cot({90}^{\circ }-{42}^{\circ }).tan({90}^{\circ }-{67}^{\circ })$
= $tan{48}^{\circ }.tan{23}^{\circ }.cot{48}^{\circ }.cot{23}^{\circ }$

We also know that, $tan\theta \times cot\theta =1$

$\therefore tan{48}^{\circ }.tan{23}^{\circ }.cot{48}^{\circ }.cot{23}^{\circ }$
= $tan{48}^{\circ }.cot{48}^{\circ }.tan{23}^{\circ }.cot{23}^{\circ }$
= $1\times 1$
= 1

ii) $cos{38}^{\circ }.cos{52}^{\circ }-sin{38}^{\circ }.sin{52}^{\circ }$

We know that, $cos\theta =sin({90}^{\circ }-\theta )$ and $cos\theta =sin({90}^{\circ }-\theta )$

$\therefore cos{38}^{\circ }.cos{52}^{\circ }-sin{38}^{\circ }.sin{52}^{\circ }$
= $cos{38}^{\circ }.sin({90}^{\circ }-{52}^{\circ })-sin{38}^{\circ }.cos({90}^{\circ }-{52}^{\circ }$
= $cos{38}^{\circ }.sin{38}^{\circ }-sin{38}^{\circ }.cos{38}^{\circ }$
= $0$