Question

If $p=cosx-sinx$, $q=\frac{1-{\sin }^{3}x}{1-\sin x},r=\frac{1+{\cos }^{3}x}{1+\cos x},$
What is the value of p + q + r ?

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is D 3

$p=cosx-sinx$, $q=\frac{1-{\sin }^{3}x}{1-\sin x},r=\frac{1+{\cos }^{3}x}{1+\cos x},$
$q=\frac{(1-\sin x)(1+si{n}^{2}x+sinx)}{1-sinx}=1+si{n}^{2}x+sinx$ ....Since, $a^3-b^3=(a-b)(a^2+ab+b^2)$
$r=\frac{(1+\cos x)(1+co{s}^{2}x-cosx)}{1+cosx}=1+co{s}^x-cosx$ ....Since, $a^3+b^3=(a+b)(a^2-ab+b^2)$
Now, $p+q+r=cosx-sinx+1+si{n}^{2}x+sinx+1+co{s}^{2}x-cosx$

= $2+si{n}^{2}x+co{s}^{2}x$ (Since, $si{n}^{2}x+co{s}^{2}x=1$)

= $3$