Vectors a- and b- include an angle - between them- If a-b- and a-b- respectively subtend angles - and - with a- then tan -tan - is -

The correct option is B (2absin θ)(a^2 b^2 cos^2 θ) Given that the angle between a⃗ and b⃗ is #952; Since (a⃗+b⃗) makes an angle of ...

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Byju's Answer

Standard XII

Physics

Vector Addition

Vectors a⃗ ...

Question

Vectors $\to a$ and $\to b$ include an angle $θ$ between them. If $(\to a + \to b)$ and $(\to a - \to b)$ respectively subtend angles $α$ and $β$ with $\to a$ , then $(t a n α + t a n β)$ is :

\frac{(a b s i n θ)}{(a^{2} + b^{2} c o s^{2} θ)}

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\frac{(2 a b s i n θ)}{(a^{2} - b^{2} c o s^{2} θ)}

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\frac{(a^{2} s i n^{2} θ)}{(a^{2} + b^{2} c o s^{2} θ)}

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\frac{(B^{2} s i n^{2} θ)}{(a^{2} - b^{2} c o s^{2} θ)}

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Solution

The correct option is B $\frac{(2 a b s i n θ)}{(a^{2} - b^{2} c o s^{2} θ)}$
Given that the angle between $\to a$ and $\to b$ is $θ$

Since $(\to a + \to b)$ makes an angle of $α$ with $\to a$ , hence

$⟹ t a n α = \frac{b s i n θ}{(a + b c o s θ)}$

Similarly, $(\to a - \to b)$ makes an angle of $β$ with $\to a$ , Hence

$⟹ t a n β = \frac{b s i n θ}{(a - b c o s θ)}$

Adding both the equation we get,

$⟹ t a n α + t a n β = \frac{b s i n θ}{(a + b c o s θ)} + \frac{b s i n θ}{(a - b c o s θ)}$

$⟹ t a n α + t a n β = b s i n θ [\frac{1}{(a + b c o s θ)} + \frac{1}{(a - b c o s θ)}]$

$⟹ t a n α + t a n β = b s i n θ [\frac{(a - b c o s θ + a + b c o s θ)}{(a + b c o s θ) (a - b c o s θ)}]$

$⟹ t a n α + t a n β = \frac{2 a b s i n θ}{a ² - b ² c o s ² θ}$

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Similar questions

Q. The vectors

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Vectors →a and →b include an angle θ between them. If (→a+→b) and (→a−→b) respectively subtend angles α and β with →a, then (tanα+tanβ) is :

Vectors $\to a$ and $\to b$ include an angle $θ$ between them. If $(\to a + \to b)$ and $(\to a - \to b)$ respectively subtend angles $α$ and $β$ with $\to a$ , then $(t a n α + t a n β)$ is :