Question

if tanθ=1/2 then sinθ/cos^3θ + cosθ/sin^3θ=??
$if \: \tan(θ) = 1 \div 2 \: then \: what \: value \: of( \: \sin(θ) \div \cos^3θ) +( \cos θ \div \sin ^3θ )$
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Answer 1

Answer:

85/8

Step-by-step explanation:

Given,

tanθ = 1/2

To Find :-

Value of :-

$\dfrac{sin\theta}{cos^3\theta}+\dfrac{cos\theta}{sin^3\theta}$

How To DO :-

As the gave value of 'tanθ' by using it we  need to find the value 'sinθ and cosθ' by using the formula and we need to substitute those values in what we need to find.

Formula Required :-

tanθ = opposite side/adjacent side

pyathgoreas theorem :-

(hypotenuse side)² = (opposite side)² + (adjacent side)²

sinθ = opposite side/hypotenuse side

cosθ = adjacent side/hypotenuse side

Solution :-

tanθ = 1/2

opposite side/adjacent side = 1/2

→ opposite side = 1 , adjacent side = 2

Let, hypotenuse side be 'x'

Applying pyathgoreas theorem :-

(hypotenuse side)² = (opposite side)² + (adjacent side)²

x² = 1² + 2²

x² = 1 + 4

x² = 5

x = √5

∴ Hypotenuse side = x = √5

sinθ = opposite side/hypotenuse side

= 1/√5

cosθ = adjacent side/hypotenuse side

= 2/√5

$\dfrac{sin\theta}{cos^3\theta}+\dfrac{cos\theta}{sin^3\theta}$

Substituting Values of 'sinθ ' and 'cosθ ' :-

$=\dfrac{\dfrac{1}{\sqrt{5}}}{\left(\dfrac{2}{\sqrt{5}}\right)^3}+\dfrac{\dfrac{2}{\sqrt{5}}}{\left(\dfrac{1}{\sqrt{5}}\right)^3}$

$=\dfrac{\dfrac{1}{\sqrt{5}}}{\dfrac{2^3}{(\sqrt{5})^3}}+\dfrac{\dfrac{2}{\sqrt{5}}}{\dfrac{1^3}{(\sqrt{5})^3}}$

[ ∴ (a/b)³ = a³/b³]

$=\dfrac{\dfrac{1}{\sqrt{5}}}{\dfrac{8}{5\sqrt{5}}}+\dfrac{\dfrac{2}{\sqrt{5}}}{\dfrac{1}{5\sqrt{5}}}$

$=\dfrac{1}{\sqrt{5}}\times \dfrac{5\sqrt{5}}{8}+\dfrac{2}{\sqrt{5}}\times \dfrac{5\sqrt{5}}{1}$

$=\dfrac{1}{1}\times \dfrac{5}{8}+\dfrac{2}{1}\times \dfrac{5}{1}$

= 5/8 + 10

Taking L.C.M :-

= 5 + 8(10)/8

= 5 + 80/8

= 85/8

$\therefore \dfrac{sin\theta}{cos^3\theta}+\dfrac{cos\theta}{sin^3\theta}=\dfrac{85}{8}$