Question

$if \sin( \alpha ) = 5 \div 13 \: \: then \: prove \: it \: that \: \tan( \alpha ) + \sec( \alpha ) = 1.5$


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Answer 1

Given :-

• sinA = (5/13) .

To Prove :-

• TanA + secA = 1.5.

Points To Remember :-

• sin theta = Perpendicular/Hypotenuse
• cos theta = Base/Hypotenuse
• tan theta = Perpendicular/Base
• cosec theta = Hypotenuse/Perpendicular
• sec theta = Hypotenuse/Base
• cot theta = Base/Perpendicular
• (Perpendicular)² + (Base)² = (Hypotenuse)²

Solution :-

→ sinA = 5/13 = Perpendicular/Hypotenuse

So,,

→ Perpendicular = 5

→ Hypotenuse = 13 .

So,

→ Base = √(H² - P²)

→ Base = √(13² - 5²)

→ Base = √(169 - 25)

→ Base = √144

→ Base = 12.

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So,

→ tan A = Perpendicular/Base

→ tanA = 5/12 .

And,

→ sec A = Hypotenuse/Base

→ secA = 13/12 .

Hence,

→ tanA + secA = (5/12) + (13/12)

→ tanA + secA = (18/12)

→ tanA + secA = 1.5 (Proved).

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Answer 2

Answer:

$given \\ \\ sin \alpha = \frac{5}{13} = \frac{p}{h} \\ \\ by \: pythagoras \: theoram \\ \\ \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \implies {(13)}^{2} = {(5)}^{2} + {b}^{2} \\ \\ \implies169 = 25 + {b}^{2} \\ \\ \implies169 - 25 = {b}^{2} \\ \\ \implies144 = {b}^{2} \\ \\ \implies b = \sqrt{144} \\ \\ \implies b = 12$

$tan \alpha = \frac{p}{b} = \frac{5}{12} \\ \\ sec \alpha = \frac{h}{b} = \frac{13}{12} \\ \\ put \: value \: of \: this \: in \: given \: equation \\ \\ \implies tan \alpha + sec \alpha = 1.5 \\ \\ \implies \frac{5}{12} + \frac{13}{12} = 1.5 \\ \\ \implies \frac{18}{12} = 1.5 \\ \\ \implies1.5 = 1.5 \\ \\ hence \: proved.$