If |a|=4, |b|=5, |a-b|=3 and theta is the ang | vedclass

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(D) Given that, $|a|=4, |b|=5, |a-b|=3$.We know that $|a-b|^2 = |a|^2 + |b|^2 - 2(a \cdot b)$.Substituting the given values:$(3)^2 = (4)^2 + (5)^2 - 2

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$\frac{9}{16}$

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(D) Given that, $|a|=4, |b|=5, |a-b|=3$.We know that $|a-b|^2 = |a|^2 + |b|^2 - 2(a \cdot b)$.Substituting the given values:$(3)^2 = (4)^2 + (5)^2 - 2(a \cdot b)$$9 = 16 + 25 - 2(a \cdot b)$$9 = 41 - 2(a \cdot b)$$2(a \cdot b) = 41 - 9 = 32$$a \cdot b = 16$.Now, the cosine of the angle $	heta$ between vectors $a$ and $b$ is given by:$\cos 	heta = \frac{a \cdot b}{|a||b|} = \frac{16}{4 	imes 5} = \frac{16}{20} = \frac{4}{5}$.Since $\cos 	heta = \frac{4}{5}$, we can form a right-angled triangle with adjacent side $4$ and hypotenuse $5$. The opposite side is $\sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$.Therefore, $	an 	heta = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{3}{4}$.Finally, $	an^2 	heta = \left(\frac{3}{4}ight)^2 = \frac{9}{16}$.

If $|a|=4, |b|=5, |a-b|=3$ and $\theta$ is the angle between the vectors $a$ and $b$, then $\tan^2 \theta=$

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