Question

If 15 $si{n}^4\alpha$ + 10 $co{s}^4\alpha$ = 6 then the value of 8 $cose{c}^6\alpha$ - 27 $se{c}^6\alpha$ is __.

Answer 1

15 $si{n}^4\alpha$ + 10 $co{s}^4\alpha$ = 6 How to proceed?

We know the identity $si{n}^2\alpha$ + $co{s}^2\alpha$ = 1

$se{c}^2\alpha$ = 1+ $ta{n}^2\alpha$

$cose{c}^2\alpha$ = 1+ $co{t}^2\alpha$

Can we use these identities here?

We can write 15 $si{n}^4\alpha$ = 5 $si{n}^4\alpha$ + 10 $si{n}^4\alpha$

5 $si{n}^4\alpha$ + 10 $si{n}^4\alpha$ + 10 $co{s}^4\alpha$ = 6

5 $si{n}^4\alpha$ + 10 ($si{n}^4\alpha$ + $co{s}^4\alpha$) = 6 -----------(1)

$(si{n}^2\alpha +co{s}^2\alpha {)}^2$ = $si{n}^4\alpha$ + $co{s}^4\alpha$ + 2 $si{n}^2\alpha$ $co{s}^2\alpha$

1 - 2 $si{n}^2\alpha$ $co{s}^2\alpha$ = $si{n}^4\alpha$ + $co{s}^4\alpha$

Substituting the value of $si{n}^4\alpha$ + $co{s}^4\alpha$ in equation 1

So, 5 $si{n}^4\alpha$ + 10 (1-2 $si{n}^2\alpha$ $co{s}^2\alpha$) = 6

5 $si{n}^4\alpha$ + 10 - 20 $si{n}^2\alpha$ (1 - $si{n}^2\alpha$) = 6

5 $si{n}^4\alpha$ + 4 - 20 $si{n}^2\alpha$ + 20 $si{n}^2\alpha$ + 20 $si{n}^4\alpha$ = 0

25$si{n}^4\alpha$ - 20 $si{n}^2\alpha$ + 4 = 0

→ Let $si{n}^2\alpha$ = x

25 $x^2$- 20x + 4 = 0 Quadratic equation

25 $x^2$ - 10x + 10x + 4 = 0

5x(5x - 2) - 2 (5x - 2) = 0

$(5x-2{)}^2$ = 0

x = $\frac{2}{5}$

$si{n}^2\alpha$ = $\frac{2}{5}$

$sin\alpha$= $\frac{\sqrt{2}}{\sqrt{5}}$

In right angle triangle

From trigonometric ratio

So, sec $\alpha$ = $\frac{\sqrt{5}}{\sqrt{2}}$

cosec $\alpha$ = $\frac{\sqrt{5}}{\sqrt{2}}$

So, 8 $cose{c}^6\alpha$ - 27 $se{c}^6\alpha$

8${\left(\frac{\sqrt{5}}{\sqrt{2}}\right)}^6$ - 27${\left(\frac{\sqrt{5}}{\sqrt{2}}\right)}^6$

= 8 $\times$ $\frac{5^3}{2^3}$ - 27 $\times$ $\frac{5^3}{3^3}$

= 8 $\times$ $\frac{125}{8}$ - 27 $\times$ $\frac{125}{27}$

= 125 - 125 = 0