Question

5 sin = 3, then find

(i) cos theeta (ii) tan theeeta (iii) sec theeta – cosec theeta​

Answer 1

Answer

sin θ=53

secθ−tanθsecθ+tanθ 

Multiplying numerator and denominator by cos θ

secθ−tanθsecθ+tanθ = 1−sinθ1+sinθ

= 1−531+53

Multiplying numerator and denominator by 5,

= 5−35+3

=4

Answer 2

Question :

If 5 sin θ = 3 then find

1) Cos θ

2)Tan θ

3)Sec θ - Cosec θ

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Answer :

Given :

$\large\orange{5 \: Sin \: θ = 3}$

$\orange{Sin \: θ = \dfrac{3}{5}=\dfrac{opposite \: to \: \angleθ}{hypotenuse \: to \: \angleθ}}$

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Refer the image now :

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1)

$\orange{Cos \: θ = \dfrac{adjacent \: to \: \angleθ}{hypotenuse \: to \: \angleθ}}$

$\red{Cos \: θ = \dfrac{4}{5}}$

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2)

$\orange{Tan \: θ = \dfrac{opposite \: to \: \angleθ}{adjacent \: to \: \angleθ}}$

$\red{Cos \: θ = \dfrac{3}{4}}$

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3)

Sec θ - Cosec θ

Now,

$\orange{sec \: θ = \dfrac{hypotenuse \: to \: \angleθ}{adjacent \: to \: \angleθ}}$

$\red{sec \: θ = \dfrac{5}{4}}$

$\orange{Cosec \: θ = \dfrac{hypotenuse \: to \: \angleθ}{opposite \: to \: \angleθ}}$

$\red{Cosec \: θ = \dfrac{5}{3}}$

Now,

Sec θ - Cosec θ

$\orange{\dfrac{5}{4}-\dfrac{5}{3}}$

$\orange{\dfrac{5×3}{4×3}-\dfrac{5×4}{3×4}}$

$\orange{\dfrac{15}{12}-\dfrac{20}{12}}$

$\orange{\dfrac{15-20}{12}}$

$\red{\dfrac{5}{12}}$

[Hope this helps you.../]