Question

Hey !!

TRIGONOMETRIC RATIOS


If Cose∅ + Cot∅ = k then prove that cos∅ = k² - 1 / k² + 1

NO COPIED ANSWERS !!​

Answer 1

$\huge \underline \mathfrak {Solution:-}$

Let ∅ = x

Cosec x + Cot x = k (given)

$\frac{ 1}{ \sin(x) } + \frac{ \cos(x) }{ \sin(x) } = k \\ \\ \frac{(1 + \cos(x) )}{ \sin(x) } = k \\ \\ \frac{2 \cos {}^{2} ( \frac{x}{2} ) }{2 \sin(x ) \cos(x) } = k \\ \\ \frac{ \cos( \frac{x}{2} ) }{ \sin( \frac{x}{2} ) } = k \: \: \: \: \: \: \: \: ......(1)$

Now, taking RHS from the question

$\frac{ {k}^{2} - 1}{ {k}^{2} + 1} \\ \\ = \frac{ \frac{ \cos {}^{2} ( \frac{x}{2} ) }{ \sin {}^{2} ( \frac{x}{2} ) } - 1 }{ \frac{ \cos {}^{2} ( \frac{x}{2} ) }{ \sin {}^{2} ( \frac{x}{2} ) } + 1 } \\ \\ = \frac{\cos {}^{2} ( \frac{x}{2} ) -\sin {}^{2} ( \frac{x}{2} ) }{\cos {}^{2} ( \frac{x}{2} ) + \sin {}^{2} ( \frac{x}{2} )} \\ \\ = \frac{ \cos(x) }{1} \\ \\ = \cos(x)$

= LHS

So, cos∅ = k² - 1 / k² + 1

Hence proved

Answer 2

Correct Question

If Cosec∅ + Cot∅ = k then prove that cos∅ = k² - 1 / k² + 1

Solution

Given : cosec∅ + cot∅ = k ___(1)

On squaring equation (1)

→ cosec²∅ + cot²∅ + 2(cosec∅)(cot∅) = k²

→ cosec²∅ + cot²∅ + 2(1/sin∅)(cos∅/sin∅)

→ 1/(sin²∅) + cos²∅/sin²∅ + 2cos∅/sin²∅

→ (1 + cos²∅ + 2cos∅)/sin²∅

→ (cos∅ + 1)²/sin²∅ = k² ___(2)

→ (1 + cos∅)(1 + cos∅)/(1 - cos²∅) = k²

→(1 + cos∅)(1 + cos∅)/(1 - cos∅)(1 + cos∅) = k²

→1 + cos∅ = k² - k²cos∅

→ k²cos∅ + cos∅ = k² - 1

→ cos∅(k² + 1) = k² - 1

→ cos∅ = (k² - 1)/(k² + 1)

Hence Proved