In a right angled triangle- basehypotenuse -

The correct option is D cosθ In a right angled triangle #160; base hypotenuse =cosθ #160; ...

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Circular Measurement of Angle

In a right an...

Question

In a right angled triangle, $\frac{b a s e}{h y p o t e n u s e} =$

sin θ

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cos θ

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tan θ

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None of these

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cos θ = \frac{1}{2} a n d sin θ = \frac{\sqrt{3}}{2}

tan θ = \sqrt{3}

tan θ = \frac{sin θ}{cos θ}

sin θ

cos θ

∠ A C B = θ .

If t a n θ = \frac{5}{12}

then find the value of \frac{(1 + c o s θ) \times (1 - s i n θ)}{(1 + s i n θ) \times (1 - c o s θ)} .

The expression $E = \frac{sec θ + tan θ - 1}{tan θ - sec θ + 1}$ can be simplified to -

\frac{h y p o t e n u s e}{b a s e} =

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