Question

10 $co{s}^{2}x$ - 6 sinxcosx + 2$si{n}^{2}x$ is equals to

• A. (A) 6 + 4 cos2x + 3sin2x
• B. (A) 6 - 4 cos2x + 3sin2x
• C. 6 + 4 cos2x - 3 sin2x
• D. 6 - 4 cos2x - 3sin2x

Answer 1

The correct option is C

6 + 4 cos2x - 3 sin2x

Option are given in terms of sin2x and cos2x.So we will use the formulae cos2x = 2$co{s}^{2}x$ - 1

and sin2x = 2sinxcosx to simplify the given expressions.

$\Rightarrow$ 10$co{s}^{2}x$ - 6sinxcosx + 2$si{n}^{2}x$

= 8$co{s}^{2}x$ - 6sinxcosx + 2 ($co{s}^{2}x$ + $si{n}^{2}x$)

= 8$\left(\frac{1+cos2x}{2}\right)$ - 3x sin2x + 2

= 4 (1 + cos2x) - 3sin2x + 2

= 6 + 4 cos2x - 3 sin2x

$\Rightarrow$ C