Question

Prove that: $\cos {510}^o\cos {330}^o+\sin {390}^o\cos {120}^o=-1$.

Answer 1

We have,

LHS

$=cos{510}^{\circ }cos{330}^{\circ }+sin{390}^{\circ }cos{120}^{\circ }$

$=cos(540-30{)}^{\circ }cos(360-30{)}^{\circ }+sin({360}^{\circ }+30{)}^{\circ }cos(90+30{)}^{\circ }$

$=(-cos{30}^{\circ })\left(cos{30}^{\circ }\right)+\left(sin{30}^{\circ }\right)\left(sin{30}^{\circ }\right)$

$q=-co{s}^2{30}^{\circ }-si{n}^2{30}^{\circ }$

$=-(co{s}^2{30}^{\circ }+si{n}^2{30}^{\circ })$

$=-1.$

$=RHS$

here, as we know,

$cos\left[\right(2n-1)\pi -\theta ]=-cos\theta$

$cos\left[\right(2n\pi )-\theta ]=cos\theta$

$sin\left[\right(2n\pi )+\theta ]=sin\theta$

$cos\left[\right(4n+1)\pi /2+\theta ]=-sin\theta$

REF .Image.

I Quadrant

All are positive.

II Quadrant

sin, cosec are positive & rest are negative

III Quadrant

only tan & cot are positive & rest are negative

IV Quadrant only

cos, sec are positive & rest are negative.