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Grade 12th passTrigonometry

Tan^3x + cot^3x =52 then what is the value of tan^2x +cot^2x

Profile image of Suryaa Natarajan
7 Years agoGrade 12th pass
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1 Answer

Profile image of Saurabh Koranglekar
7 Years ago

To solve the equation \( \tan^3 x + \cot^3 x = 52 \) and find the value of \( \tan^2 x + \cot^2 x \), we can leverage some algebraic identities and properties of tangent and cotangent functions. Let's break it down step by step.

Using Algebraic Identities

First, let's recall the identity for the sum of cubes. The formula for \( a^3 + b^3 \) can be expressed as:

  • a^3 + b^3 = (a + b)(a^2 - ab + b^2)

In this case, let \( a = \tan x \) and \( b = \cot x \). Thus, we can rewrite our equation:

So, \( \tan^3 x + \cot^3 x = (\tan x + \cot x)(\tan^2 x - \tan x \cdot \cot x + \cot^2 x) \).

Expressing in Terms of Tangent

We also know that \( \tan x \cdot \cot x = 1 \) (since they are reciprocal functions). Therefore, we can simplify the expression:

  • \(\tan^2 x - 1 + \cot^2 x\)

This means we can replace \( \cot^2 x \) with \( \frac{1}{\tan^2 x} \), if needed, when we go deeper.

Defining Variables

Let \( t = \tan x \). Then \( \cot x = \frac{1}{t} \). We can express the original equation as:

  • t^3 + \frac{1}{t^3} = 52

Finding a Common Expression

To combine \( t^3 \) and \( \frac{1}{t^3} \), we can utilize the identity that relates \( t + \frac{1}{t} \) to \( t^3 + \frac{1}{t^3} \):

If we let \( u = t + \frac{1}{t} \), then we have:

  • t^3 + \frac{1}{t^3} = u^3 - 3u

Substituting this back into our original equation gives:

  • u^3 - 3u = 52

Solving the Cubic Equation

This is now a cubic equation which can be rearranged to:

  • u^3 - 3u - 52 = 0

Finding \( u \) may involve some trial and error or numerical methods. By checking for potential integer solutions, we can try \( u = 4 \):

  • 4^3 - 3(4) - 52 = 64 - 12 - 52 = 0

Thus, \( u = 4 \) is indeed a solution.

Calculating the Required Sum

Now that we have \( u = t + \frac{1}{t} = 4 \), we can find \( t^2 + \frac{1}{t^2} \) using the identity:

  • t^2 + \frac{1}{t^2} = (t + \frac{1}{t})^2 - 2 = u^2 - 2

Substituting in our value for \( u \):

  • t^2 + \frac{1}{t^2} = 4^2 - 2 = 16 - 2 = 14

Final Result

Therefore, the value of \( \tan^2 x + \cot^2 x \) is:

14

This approach illustrates how we can manipulate functions and their properties to derive new values effectively. Let me know if you have any further questions or if you want to dive into another topic!