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2 years ago

If tanA=a then find sin4A-2sin2A/ sin4A+2sin2A

Mathematics

1 answer:

anygoal [31]2 years ago

6 0

Answer:

The value of the given expression is

$\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2$

Step by step Explanation:

Given that $tanA=a$

To find the value of $\frac{sin4A-2sin2A}{sin4A+2sin2A}$

Let us find the value of the expression :

$\frac{sin4A-2sin2A}{sin4A+2sin2A}=\frac{2cos2Asin2A-2sin2A}{2cos2Asin2A+2sin2A}$ ( by using the formula $sin2A=2cosAsinA$ here A=2A)

$=\frac{2sin2A(cos2A-1)}{2sin2A(cos2A+1)}$

$=\frac{(cos2A-1)}{(cos2A+1)}$

$=\frac{(-(1-cos2A))}{(1+cos2A)}$ (using $sin^2A+cos^2A=1$ here A=2A)

$=\frac{-(sin^2A+cos^2A-(cos^2A-sin^2A))}{sin^2A+cos^2A+(cos^2A-sin^2A)}$ (using $cos2A=cos^2A-sin^2A$ here A=2A)

$=\frac{-(sin^2A+cos^2A-cos^2A+sin^2A)}{sin^2A+cos^2A+(cos^2A-sin^2A)}$

$=\frac{-(sin^2A+sin^2A)}{cos^2A+cos^2A}$

$=\frac{-2sin^2A}{2cos^2A}$

$=-\frac{sin^2A}{cos^2A}$

$=-tan^2A$ ( using $tanA=\frac{sinA}{cosA}$ here A=2A )

$=-a^2$ (since tanA=a given )

Therefore $\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2$

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2 years ago

Find the base of an isosceles triangle whose area is 60 cm² and the length of one of its equal sides is 13 cm.

Kobotan [32]

Answer:

Base = 24 cm or 10cm

Step-by-step explanation:

REMEMBER:
An isosceles triangle ABC with base BC = ‘b' & height AD = ‘h' & its equal sides =13 cm & area = 60 cm²

Using the formulas

$A=\frac{bh_b}{2}$

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There are 2 solutions for $b$

$b=2\sqrt{2} \frac{A}{\sqrt{a^2+\sqrt{a^4-4A^2} } } =2*\sqrt{2} *\frac{60}{\sqrt{13^2+\sqrt{13^4-4*60^2} } }$ ≈ $10cm$ $b=2\sqrt{2} \frac{A}{\sqrt{a^2+\sqrt{a^4-4A^2} } } =2*\sqrt{2} *\frac{60}{\sqrt{13^2-\sqrt{13^4-4*60^2} } }=24cm$

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Area of a triangle = 1/2 * b * h = 60

=> h = 120/b

In right triangle ABD

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=>169 = 120²/b² + b²/4

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=> b^4 - 676 b² + 57600 = 0

=> b² = 676 +- √(676² - 4*57600) / 2

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=> b² = (676 +- 476 )/2

=> b² = 1152/2 , 200 /2

=> b² = 576 , 100

=> b = 24, 10

So, Base = 24 cm or 10cm

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If tanA=a then find sin4A-2sin2A/ sin4A+2sin2A​

If tanA=a then find sin4A-2sin2A/ sin4A+2sin2A