Question

$\frac{cos \: 45 + 2cos \: 90}{sin \: 45 + 2sin \: 90}$
evaluate​

Answer 1

CORRECT QUESTION :-

$\\ \implies \sf \: evaluate \: \: \dfrac{ \cos45 ^{ \circ} + 2 \cos90 ^{ \circ} }{ \sin45 ^{ \circ} + 2 \sin90 ^{ \circ} }$

GIVEN :-

$\\ \implies \sf \: \dfrac{ \cos45 ^{ \circ} + 2 \cos90 ^{ \circ} }{ \sin45 ^{ \circ} + 2 \sin90 ^{ \circ} }$

TO FIND :-

$\\ \implies \sf \: value \: of \: \: \dfrac{ \cos45 ^{ \circ} + 2 \cos90 ^{ \circ} }{ \sin45 ^{ \circ} + 2 \sin90 ^{ \circ} }$

SOLUTION :-

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 65^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

★ Refer to the trigonometric ratios table for the required values.

$\implies \sf \: \dfrac{ \cos45 ^{ \circ} + 2 \cos90 ^{ \circ} }{ \sin45 ^{ \circ} + 2 \sin90 ^{ \circ} } \\ \\ \\ \implies \sf\dfrac{ \left(\dfrac{1 }{ \sqrt{2} } + 2 \times 0 \right)}{ \left(\dfrac{1}{ \sqrt{2} } \: + 2 \times 1 \right)} \\ \\ \\ \implies \sf \: \dfrac{ \left(\dfrac{1}{\sqrt{2}} \right)}{ \left(\dfrac{1+2\sqrt{2}}{\sqrt{2}} \right)} \\ \\ \implies \sf \dfrac{1}{ \cancel{\sqrt{2}}} \times \dfrac{\sqrt{2}}{1 + 2\ \cancel{\sqrt{2}}} \\ \\ \\ \implies \sf \dfrac{1}{1 + 2\sqrt{2}} \\ \\ \\ \implies \underline{ \boxed{ \red{ \sf \: \dfrac{ \cos45 ^{ \circ} + 2 \cos90 ^{ \circ} }{ \sin45 ^{ \circ} + 2 \sin90 ^{ \circ} } = \dfrac{1}{1 + 2\sqrt{2}}}}}$

∴ The required answer is $\dfrac{1}{1 + 2\sqrt{2}}$.

Answer 2

Answer:

1/1+2√2

Step-by-step explanation:

=> cos45°+2cos90° / sin45°+2sin90°

=> (1/√2 + 2*0) / (1/√2 + 2*1)

=> (1/√2) / (1+2√2 / √2)

=> 1/√2 × √2 / 1+2√2

=> 1/1+2√2

Therefore, Required value is 1/1+2√2

hope it helps !! ( ◜‿◝ )♡