Question

please please please solve this math
$if \: \sin( \alpha ) = 5 \div 13 \: then \: proved \: that \: \tan( \alpha ) + \sec( \alpha ) = 1.5$
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Answer 1

We know that, according to a right triangle, the sine ratio of an acute angle in the triangle is the ratio of the side opposite to the angle to the hypotenuse.

$\sin\theta=\dfrac{opposite\ side}{hypotenuse}$

Here the sine ratio of an arbitrary angle α is given as 5/13.

Thus, we have to assume a right triangle with an angle α, let be acute, hypotenuse of length 13 units and the side opposite to the angle α of length 5 units.

What will be the length of the side adjacent to the angle α?

We can find it by using Pythagoras' theorem.

√(13² - 5²) = √(169 - 25) = √144 = 12 units.

Thus,

$\tan\alpha=\dfrac{opposite\ side}{adjacent\ side}=\dfrac{5}{12}\\ \\ \sec\alpha=\dfrac{hypotenuse}{adjacent \ side}=\dfrac{13}{12}\\ \\ \\ \text{Now},\ \ \tan\alpha + \sec\alpha=\dfrac{5}{12}+\dfrac{13}{12}=\dfrac{18}{12}=1.5$

Hence Proved!!!

Answer 2

hope it helps!

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