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Q.

If $\tan θ = \frac{a}{b}$ , then $\cos θ =$

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a

$\frac{a}{\sqrt{a^{2} + b^{2}}}$

b

$\frac{a}{a^{2} + b^{2}}$

c

$\frac{b}{\sqrt{a^{2} + b^{2}}}$

d

$\frac{b}{a^{2} + b^{2}}$

answer is D.

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Detailed Solution

$\tan θ = \frac{a}{b}$

Question Image

$A B^{2} = A C^{2} + B C^{2}$

$A B^{2} = a^{2} + b^{2}$

$\cos θ = \frac{a d j a c e n t t o θ}{h y p o t e n u s e}$

$\cos θ = \frac{b}{\sqrt{a^{2} + b^{2}}}$

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If tanθ=ab, then cosθ=