If `15 sin^(4)alpha+10cos^(4)alpha=6`, then the value of `8 cosec^(6)alpha+27 sec^(6)alpha` is

Options

A. 150
B. 175
C. 225
D. 250

Explanation

To solve the problem, we start with the equation given:

**Step 1: Start with the equation**
\[ 15 \sin^4 \alpha + 10 \cos^4 \alpha = 6 \]

**Step 2: Divide both sides by \( \cos^4 \alpha \)**
\[ \frac{15 \sin^4 \alpha}{\cos^4 \alpha} + \frac{10 \cos^4 \alpha}{\cos^4 \alpha} = \frac{6}{\cos^4 \alpha} \]
This simplifies to:
\[ 15 \tan^4 \alpha + 10 = 6 \sec^4 \alpha \]

**Step 3: Rewrite \( \sec^4 \alpha \) using the identity**
We know that:
\[ \sec^2 \alpha = 1 + \tan^2 \alpha \]
Thus,
\[ \sec^4 \alpha = (1 + \tan^2 \alpha)^2 = 1 + 2 \tan^2 \alpha + \tan^4 \alpha \]

**Step 4: Substitute \( \sec^4 \alpha \) in the equation**
Substituting this into our equation gives:
\[ 15 \tan^4 \alpha + 10 = 6(1 + 2 \tan^2 \alpha + \tan^4 \alpha) \]

**Step 5: Expand and rearrange the equation**
Expanding the right side:
\[ 15 \tan^4 \alpha + 10 = 6 + 12 \tan^2 \alpha + 6 \tan^4 \alpha \]
Rearranging gives:
\[ 15 \tan^4 \alpha - 6 \tan^4 \alpha - 12 \tan^2 \alpha + 10 - 6 = 0 \]
This simplifies to:
\[ 9 \tan^4 \alpha - 12 \tan^2 \alpha + 4 = 0 \]

**Step 6: Let \( x = \tan^2 \alpha \)**
Let \( x = \tan^2 \alpha \):
\[ 9x^2 - 12x + 4 = 0 \]

**Step 7: Solve the quadratic equation**
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 9, b = -12, c = 4 \):
\[ x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 9 \cdot 4}}{2 \cdot 9} \]
\[ x = \frac{12 \pm \sqrt{144 - 144}}{18} \]
\[ x = \frac{12}{18} = \frac{2}{3} \]

Thus,
\[ \tan^2 \alpha = \frac{2}{3} \]

**Step 8: Find \( \sec^2 \alpha \) and \( \csc^2 \alpha \)**
Using the identity \( \sec^2 \alpha = 1 + \tan^2 \alpha \):
\[ \sec^2 \alpha = 1 + \frac{2}{3} = \frac{5}{3} \]
Thus,
\[ \sec^6 \alpha = \left(\frac{5}{3}\right)^3 = \frac{125}{27} \]

For \( \csc^2 \alpha \):
Using \( \csc^2 \alpha = 1 + \cot^2 \alpha \) and \( \cot^2 \alpha = \frac{1}{\tan^2 \alpha} = \frac{3}{2} \):
\[ \csc^2 \alpha = 1 + \frac{3}{2} = \frac{5}{2} \]
Thus,
\[ \csc^6 \alpha = \left(\frac{5}{2}\right)^3 = \frac{125}{8} \]

**Step 9: Calculate \( 8 \csc^6 \alpha + 27 \sec^6 \alpha \)**
Now, substituting these values:
\[ 8 \csc^6 \alpha + 27 \sec^6 \alpha = 8 \left(\frac{125}{8}\right) + 27 \left(\frac{125}{27}\right) \]
\[ = 125 + 125 = 250 \]

Thus, the final answer is:
\[ \boxed{250} \]

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