Paragraph: It is given that (θ-α)=a, (θ-β)=b Question: What is ^2(α-β)+2 a b (α-β) equal to?

Correct answer: A. a^2+b^2 ^2(α-β)+2 ab (α-β)=1- ^2(α-β)+2 ab aligned & (α-β)=1- (α-β)[ (α-β)-2 ab] & = 1-(a b+ 1-a^2 1-b^2 )[a b+ 1-a^2 1-b^2 -2 a b] &= 1-( 1-a^2 1-b^2 +a b)( 1-a^2 1-b^2 -a b) &= 1-[( 1-a^2 1-b^2 )^2-(a b)^2] aligned aligned & =1-[(1-a^2)(1-b^2)-a^2 b^2) & =1-(1-b^2-a^2+a^2 b^2-a^2 b^2) & =1-1+b^2+a^2=a^2+b^2 aligned

NDA 2019

Exam: NDA
Year: 2019
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

NDA 2019 · Mathematics · Trigonometric Ratios & Identities

Paragraph: It is given that (θ-α)=a, (θ-β)=b Question: What is...

Paragraph:It is given that

\cos (\theta-\alpha)=a, \cos (\theta-\beta)=b

Question:What is

\sin ^2(\alpha-\beta)+2 a b \cos (\alpha-\beta)

equal to?

Options

A $a^2+b^2$
B $a^2-b^2$
C $\mathrm{b}^2-\mathrm{a}^2$
D $- \left(a²+b²\right)$

Correct answer

A. $a^2+b^2$

Step-by-step solution

\sin ^2(\alpha-\beta)+2 \mathrm{ab} \cos (\alpha-\beta)=1-\cos ^2(\alpha-\beta)+2 \mathrm{ab}

\begin{aligned} & \cos (\alpha-\beta)=1-\cos (\alpha-\beta)[\cos (\alpha-\beta)-2 \mathrm{ab}] \\ & \quad= 1-\left(a b+\sqrt{1-a^2} \sqrt{1-b^2}\right)\left[a b+\sqrt{1-a^2} \sqrt{1-b^2}-2 a b\right] \\ &= 1-\left(\sqrt{1-a^2} \sqrt{1-b^2}+a b\right)\left(\sqrt{1-a^2} \sqrt{1-b^2}-a b\right) \\ &= 1-\left[\left(\sqrt{1-a^2} \sqrt{1-b^2}\right)^2-(a b)^2\right]\end{aligned}

$\begin{aligned} & =1-\left[\left(1-a^2\right)\left(1-b^2\right)-a^2 b^2\right) \\ & =1-\left(1-b^2-a^2+a^2 b^2-a^2 b^2\right) \\ & =1-1+b^2+a^2=a^2+b^2\end{aligned}

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