Evaluate the following- sin 41-cos 49-

Given, #160; sin 41^∘cos 49^∘ ⇒ #160; sin(90^∘ 49^∘)cos 49^∘ We know that, sin(90^∘ θ) = cos θ ∴ cos 49^∘co ...

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Standard X

Mathematics

Complementary Trigonometric Ratios

Evaluate the ...

Question

Evaluate the following:
$\frac{sin 41^{\circ}}{cos 49^{\circ}}$

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Solution

Given, $\frac{s i n 41^{\circ}}{c o s 49^{\circ}}$

$\Rightarrow \frac{s i n (90^{\circ} - 49^{\circ})}{c o s 49^{\circ}}$

We know that, $s i n (90^{\circ} - θ) = c o s θ$

$∴ \frac{c o s 49^{\circ}}{c o s 49^{\circ}} = 1$

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, then

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show that sin 19 ° + sin 41 ° + sin 83 °= sin 23 ° + Sin 37 °+ sin 79 °

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Evaluate the following:sin41∘cos49∘

Given, sin41∘cos49∘⇒sin(90∘−49∘)cos49∘We know that, sin(90∘−θ)=cosθ∴ cos49∘cos49∘=1

Evaluate the following:
$\frac{sin 41^{\circ}}{cos 49^{\circ}}$

Given, $\frac{s i n 41^{\circ}}{c o s 49^{\circ}}$

$\Rightarrow \frac{s i n (90^{\circ} - 49^{\circ})}{c o s 49^{\circ}}$

We know that, $s i n (90^{\circ} - θ) = c o s θ$

$∴ \frac{c o s 49^{\circ}}{c o s 49^{\circ}} = 1$