MHTCET 13 Apr 2026 Shift 2 Paper

Concept:Use substitution of angles and standard trigonometric identities to reduce the expression.Explanation:Let $x=\theta-\alpha$ and $y=\theta-\beta$.Then $\alpha-\beta = y-x$ because $y-x = (\theta-\beta)-(\theta-\alpha) = \alpha-\beta$.Given $\cos x=a$ and $\sin y=b$.The required expression becomes $\cos^2(y-x)+2ab\sin(y-x)+\cos^2 x$.Since $a=\cos x$ and $b=\sin y$, write it as $\cos^2(y-x)+2\cos x\sin y\sin(y-x)+\cos^2 x$.Use $\cos(y-x)=\cos y\cos x+\sin y\sin x = a\cos y + b\sin x$.Also $\sin(y-x)=\sin y\cos x-\cos y\sin x = ab-\cos y\sin x$.Square the first identity: $\cos^2(y-x)=a^2\cos^2 y+2ab\cos y\sin x+b^2\sin^2 x$.Now $2ab\sin(y-x)=2ab(ab-\cos y\sin x)=2a^2b^2-2ab\cos y\sin x$.Add all terms, including $\cos^2 x=a^2$.The terms $2ab\cos y\sin x$ and $-2ab\cos y\sin x$ cancel each other.The remaining expression is $a^2\cos^2 y+b^2\sin^2 x+2a^2b^2+a^2$.Substitute $\cos^2 y=1-\sin^2 y=1-b^2$ and $\sin^2 x=1-\cos^2 x=1-a^2$.So we get $a^2(1-b^2)+b^2(1-a^2)+2a^2b^2+a^2$.Simplify: $a^2-a^2b^2+b^2-a^2b^2+2a^2b^2+a^2=2a^2+b^2$.Answer:The value is $2a^2+b^2$, which is option D.