If tan - -8-15 then cosec - - - a 17-8 b 8-17 c 17-15 d 15-17

Tan A= 8/15 = perpendicular/base ⇒ perpendicular = 8x, base =15x ∴ hypotenuese =√((15x)^2+(8x)^2) = √(225x^2+64x^2) = √(289x^2) =17x ⇒ cosec A= hypotenuese/ pe ...

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Standard X

Mathematics

Relation between Trigonometric Ratios

If tan θ =8/1...

Question

If tan $θ$ = $\frac{8}{15}$ then cosec $θ$ = ?

(a) $\frac{17}{8}$ (b) $\frac{8}{17}$ (c) $\frac{17}{15}$ (d) $\frac{15}{17}$

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Solution

$t a n A = \frac{8}{15} = \frac{p e r p e n d i c u l a r}{b a s e}$

$\Rightarrow p e r p e n d i c u l a r = 8 x, b a s e = 15 x$

$∴ h y p o t e n u e s e = \sqrt{(15 x)^{2} + (8 x)^{2}} = \sqrt{225 x^{2} + 64 x^{2}} = \sqrt{289 x^{2}} = 17 x$

$\Rightarrow c o s e c A = \frac{h y p o t e n u e s e}{p e r p e n d i c u l a r} = \frac{17 x}{8 x} = \frac{17}{8}$

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If tan θ =815 then cosec θ = ? (a) 178 (b) 817 (c) 1715 (d) 1517

tanA=815=perpendicularbase ⇒perpendicular=8x,base=15x ∴hypotenuese=√(15x)2+(8x)2=√225x2+64x2=√289x2=17x ⇒cosecA=hypotenueseperpendicular=17x8x=178

If tan $θ$ = $\frac{8}{15}$ then cosec $θ$ = ?

(a) $\frac{17}{8}$ (b) $\frac{8}{17}$ (c) $\frac{17}{15}$ (d) $\frac{15}{17}$