If $15 \sin^4 \alpha + 10 \cos^4 \alpha = 6$, for some $\alpha \in R$, then the value of $27 \sec^6 \alpha + 8 \cosec^6 \alpha$ is equal to: Options: (A) 350 (B) 500 (C) 400 (D) 250

### Analyzing the Setup The given equation is $15 \sin^4 \alpha + 10 \cos^4 \alpha = 6$. To simplify this expression, we divide the entire equation by $\cos^4 \alpha$. This transformation yields: $$15 \tan^4 \alpha + 10 = 6 \sec^4 \alpha$$ ### The Master Equation We utilize the fundamental trigonometric identity $\sec^2 \alpha = 1 + \tan^2 \alpha$. Substituting this into our equation, we obtain: $$15 \tan^4 \alpha + 10 = 6(1 + \tan^2 \alpha)^2$$ Expanding the right side of the equation: $$15 \tan^4 \alpha + 10 = 6(1 + 2 \tan^2 \alpha + \tan^4 \alpha)$$ Distributing the constant $6$ and rearranging the terms to one side: $$15 \tan^4 \alpha + 10 = 6 + 12 \tan^2 \alpha + 6 \tan^4 \alpha$$ $$9 \tan^4 \alpha - 12 \tan^2 \alpha + 4 = 0$$ ### Solving for the Variable The resulting quadratic in terms of $\tan^2 \alpha$ is a perfect square: $$(3 \tan^2 \alpha - 2)^2 = 0$$ This implies that: $$\tan^2 \alpha = \frac{2}{3}$$ From this result, we can derive the values for $\sin^2 \alpha$ and $\cos^2 \alpha$: $$\sin^2 \alpha = \frac{2}{5}, \quad \cos^2 \alpha = \frac{3}{5}$$ ### Final Calculation We now evaluate the target expression $27 \sec^6 \alpha + 8 \csc^6 \alpha$. Substituting the values $\sec^2 \alpha = \frac{5}{3}$ and $\csc^2 \alpha = \frac{5}{2}$: $$27 \left( \frac{5}{3} \right)^3 + 8 \left( \frac{5}{2} \right)^3$$ $$27 \left( \frac{125}{27} \right) + 8 \left( \frac{125}{8} \right)$$ $$125 + 125 = 250$$ The final value of the expression is **250**.

JEE Main 2021 (March) (18 March Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: If $15 sin^{4} α + 10 cos^{4} α = 6$ , for some $α \in R$ , then the value of $27 sec^{6} α + 8 cosec^{6} α$ is equal to:

Select Answer:

Visualized Solution

Analyze the Given Equation

Given equation: $15 sin^{4} α + 10 cos^{4} α = 6$
Target expression: $27 sec^{6} α + 8 csc^{6} α$
Objective: Solve for $α$ or find the values of $sin^{2} α$ and $cos^{2} α$ .

Divide by $cos^{4} α$

Divide the entire equation by $cos^{4} α$ :
$\frac{15 s i n ^{4} α}{c o s ^{4} α} + \frac{10 c o s ^{4} α}{c o s ^{4} α} = \frac{6}{c o s ^{4} α}$
$15 tan^{4} α + 10 = 6 sec^{4} α$

Apply Trigonometric Identity

Use the identity: $sec^{2} α = 1 + tan^{2} α$
Substitute $sec^{4} α = (1 + tan^{2} α)^{2}$ into the equation:
$15 tan^{4} α + 10 = 6 (1 + tan^{2} α)^{2}$

Expand the Algebraic Expression

Expand $(1 + tan^{2} α)^{2}$ using $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ :
$15 tan^{4} α + 10 = 6 (1 + 2 tan^{2} α + tan^{4} α)$

Distribute and Rearrange

Distribute the constant $6$ :
$15 tan^{4} α + 10 = 6 + 12 tan^{2} α + 6 tan^{4} α$
Rearrange all terms to one side:
$(15 - 6) tan^{4} α - 12 tan^{2} α + (10 - 6) = 0$
$9 tan^{4} α - 12 tan^{2} α + 4 = 0$

Recognize the Perfect Square

Observe the quadratic form in $tan^{2} α$ :
$(3 tan^{2} α)^{2} - 2 (3 tan^{2} α) (2) + (2)^{2} = 0$
This is a perfect square: $(3 tan^{2} α - 2)^{2} = 0$

Solve for $tan^{2} α$

Set the base of the square to zero:
$3 tan^{2} α - 2 = 0$
Solve for $tan^{2} α$ :
$tan^{2} α = \frac{2}{3}$

Find $sin^{2} α$ and $cos^{2} α$

Since $tan^{2} α = \frac{2}{3}$ , we can write:
$sin^{2} α = \frac{t a n ^{2} α}{1 + t a n ^{2} α} = \frac{\frac{2}{3}}{1 + \frac{2}{3}} = \frac{2}{5}$
$cos^{2} α = \frac{1}{1 + t a n ^{2} α} = \frac{1}{1 + \frac{2}{3}} = \frac{3}{5}$

Analyze the Target Expression

Target: $27 sec^{6} α + 8 csc^{6} α$
Rewrite in terms of $cos^{2} α$ and $sin^{2} α$ :
$= 27 (\frac{1}{c o s ^{2} α})^{3} + 8 (\frac{1}{s i n ^{2} α})^{3}$

Substitute the Values

Substitute $cos^{2} α = \frac{3}{5}$ and $sin^{2} α = \frac{2}{5}$ :
$= 27 (\frac{5}{3})^{3} + 8 (\frac{5}{2})^{3}$

Calculate the Cubes

Evaluate the cubes:
$= 27 (\frac{125}{27}) + 8 (\frac{125}{8})$

Final Summation

Cancel the common factors:
$= 125 + 125$
Final result: $250$

00:00 / 00:00

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The given equation is

15 sin^{4} α + 10 cos^{4} α = 6

. To simplify this expression, we divide the entire equation by

cos^{4} α

This transformation yields:

15 tan^{4} α + 10 = 6 sec^{4} α

The Master Equation

We utilize the fundamental trigonometric identity

sec^{2} α = 1 + tan^{2} α

. Substituting this into our equation, we obtain:

15 tan^{4} α + 10 = 6 (1 + tan^{2} α)^{2}

Expanding the right side of the equation:

15 tan^{4} α + 10 = 6 (1 + 2 tan^{2} α + tan^{4} α)

Distributing the constant

6

and rearranging the terms to one side:

15 tan^{4} α + 10 = 6 + 12 tan^{2} α + 6 tan^{4} α

9 tan^{4} α - 12 tan^{2} α + 4 = 0

Solving for the Variable

The resulting quadratic in terms of

tan^{2} α

is a perfect square:

(3 tan^{2} α - 2)^{2} = 0

This implies that:

tan^{2} α = \frac{2}{3}

From this result, we can derive the values for

sin^{2} α

and

cos^{2} α

sin^{2} α = \frac{2}{5}, cos^{2} α = \frac{3}{5}

Final Calculation

We now evaluate the target expression

27 sec^{6} α + 8 csc^{6} α

. Substituting the values

sec^{2} α = \frac{5}{3}

and

csc^{2} α = \frac{5}{2}

27 (\frac{5}{3})^{3} + 8 (\frac{5}{2})^{3}

27 (\frac{125}{27}) + 8 (\frac{125}{8})

125 + 125 = 250

The final value of the expression is 250.

Common Traps & FAQs

Why did we divide by

cos^{4} α

? Couldn't we have divided by

sin^{4} α

That is a fantastic question! You absolutely could have divided by

sin^{4} α

. That would have given you an equation in terms of

cot α

. It is a perfectly valid path, and you would arrive at the same final result. We chose

tan α

simply because it is the most common convention in trigonometry, but never feel restricted by the path taken in the solution—mathematics is about exploration!

Is it always safe to assume

cos α \neq = 0

when dividing?

You are thinking like an engineer! Before dividing by any term, we must ensure it is not zero. If

cos α = 0

, then

sin α

must be

1

- 1

. Plugging that into the original equation gives

15 (1)^{4} + 10 (0)^{4} = 15

, which is not

6

. Since the equation holds true only for specific values, and

cos α = 0

is not one of them, we are perfectly safe to divide.

Why did the numbers cancel out so perfectly at the end?

That is the hallmark of a well-crafted JEE problem! The examiners design these questions so that the intermediate values (like

sin^{2} α = 2/5

) are 'clean' fractions. When you see powers like

6

in the target expression, it is a subtle hint that the base values will likely be simple fractions that cube nicely. Trust the process!

Animated Solution for Mathematics - Trigonometry: If $15 sin^{4} α + 10 cos^{4} α = 6$ , for some $α \in R$ , then the value of $27 sec^{6} α + 8 cosec^{6} α$ is equal to:

Select Answer:

Visualized Solution

Analyze the Given Equation

Divide by $cos^{4} α$

Apply Trigonometric Identity

Expand the Algebraic Expression

Distribute and Rearrange

Recognize the Perfect Square

Solve for $tan^{2} α$

Find $sin^{2} α$ and $cos^{2} α$

Analyze the Target Expression

Substitute the Values

Calculate the Cubes

Final Summation

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Master Equation

Solving for the Variable

Final Calculation

Similar Questions

If $10 sin^{4} θ + 15 cos^{4} θ = 6$ , then the value of $\frac{27 cosec ^{6} θ + 8 s e c ^{6} θ}{16 s e c ^{8} θ}$ is:

If $10 sin^{4} θ + 15 cos^{4} θ = 6$ , then the value of $\frac{27 cosec ^{6} θ + 8 s e c ^{6} θ}{16 s e c ^{8} θ}$ is:

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 15sin4α+10cos4α=6, for some α∈R, then the value of 27sec6α+8cosec6α is equal to:

Select Answer:

Visualized Solution

Analyze the Given Equation

Divide by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos4α

Apply Trigonometric Identity

Expand the Algebraic Expression

Distribute and Rearrange

Recognize the Perfect Square

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan2α

Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos2α

Analyze the Target Expression

Substitute the Values

Calculate the Cubes

Final Summation

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Master Equation

Solving for the Variable

Final Calculation

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Animated Solution for Mathematics - Trigonometry: If $15 sin^{4} α + 10 cos^{4} α = 6$ , for some $α \in R$ , then the value of $27 sec^{6} α + 8 cosec^{6} α$ is equal to:

Divide by $cos^{4} α$

Solve for $tan^{2} α$

Find $sin^{2} α$ and $cos^{2} α$

If $10 sin^{4} θ + 15 cos^{4} θ = 6$ , then the value of $\frac{27 cosec ^{6} θ + 8 s e c ^{6} θ}{16 s e c ^{8} θ}$ is:

If $10 sin^{4} θ + 15 cos^{4} θ = 6$ , then the value of $\frac{27 cosec ^{6} θ + 8 s e c ^{6} θ}{16 s e c ^{8} θ}$ is:

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The value of $tan 9^{\circ} - tan 2 7^{\circ} - tan 6 3^{\circ} + tan 8 1^{\circ}$ is _____.

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