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Question

If cosecx + cotx = 2, then cosecx = ?

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is

1.25

Let's solve the given trigonometric equation to find the value of $\text{cosec}x$. We are given the equation:

$\text{cosec}x + \text{cot}x = 2$

We need to find the value of $\text{cosec}x$. To do this, we can use a fundamental trigonometric identity that relates $\text{cosec}x$ and $\text{cot}x$. The identity is:

$\text{cosec}^2x - \text{cot}^2x = 1$

This identity is in the form of a difference of squares, $a^2 - b^2 = (a-b)(a+b)$. Applying this to the identity, we get:

$(\text{cosec}x - \text{cot}x)(\text{cosec}x + \text{cot}x) = 1$

We already know from the problem statement that $\text{cosec}x + \text{cot}x = 2$. We can substitute this value into the factored identity:

$(\text{cosec}x - \text{cot}x)(2) = 1$

Now, we can solve for the term $(\text{cosec}x - \text{cot}x)$:

$\text{cosec}x - \text{cot}x = \frac{1}{2}$

So, we now have a system of two linear equations involving $\text{cosec}x$ and $\text{cot}x$:

  • Equation 1: $\text{cosec}x + \text{cot}x = 2$
  • Equation 2: $\text{cosec}x - \text{cot}x = \frac{1}{2}$

To find $\text{cosec}x$, we can add these two equations together. Notice that the $\text{cot}x$ term in the first equation is positive, and in the second equation, it's negative. When we add them, they will cancel out.

Add (Equation 1) and (Equation 2):

$(\text{cosec}x + \text{cot}x) + (\text{cosec}x - \text{cot}x) = 2 + \frac{1}{2}$

Combine like terms:

$\text{cosec}x + \text{cosec}x + \text{cot}x - \text{cot}x = 2 + 0.5$

$2 \times \text{cosec}x = 2.5$

Now, solve for $\text{cosec}x$ by dividing both sides by 2:

$\text{cosec}x = \frac{2.5}{2}$

$\text{cosec}x = 1.25$

Thus, the value of $\text{cosec}x$ is 1.25.

Revision Table: Trigonometry Identities and Solving Equations

Concept Identity/Method Application Here
Pythagorean Identity $\text{cosec}^2\theta - \text{cot}^2\theta = 1$ Used to find a second equation involving $\text{cosec}x$ and $\text{cot}x$.
Difference of Squares $a^2 - b^2 = (a-b)(a+b)$ Factored $\text{cosec}^2x - \text{cot}^2x$.
Solving System of Linear Equations Addition or Substitution Added two equations ($\text{cosec}x + \text{cot}x = 2$ and $\text{cosec}x - \text{cot}x = 0.5$) to eliminate $\text{cot}x$.

Additional Information: Understanding Cosecant and Cotangent

The trigonometric functions $\text{cosec}x$ and $\text{cot}x$ are reciprocals and ratios related to sine and tangent:

  • The cosecant function, $\text{cosec}x$, is the reciprocal of the sine function, i.e., $\text{cosec}x = \frac{1}{\text{sin}x}$. It is defined for all real numbers except where $\text{sin}x = 0$ (i.e., $x = n\pi$ for any integer $n$).
  • The cotangent function, $\text{cot}x$, is the reciprocal of the tangent function, i.e., $\text{cot}x = \frac{1}{\text{tan}x} = \frac{\text{cos}x}{\text{sin}x}$. It is also defined for all real numbers except where $\text{sin}x = 0$.
  • The identity $\text{cosec}^2x - \text{cot}^2x = 1$ is derived from the Pythagorean identity $\text{sin}^2x + \text{cos}^2x = 1$ by dividing all terms by $\text{sin}^2x$.

Problems like this one often require recognizing and applying these fundamental identities to simplify expressions or solve equations involving trigonometric functions.

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