Question

(Cosec theta-sin theta) (sec theta-cos theta)=1/tan theta plus cot theta

Answer 1

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Answer 2

Complete Question

Prove

(cosecθ - sinθ)(secθ - cosθ) = 1/(tanθ + cotθ)

Answer

Therefore, LHS = RHS. Hence proved.

Given

(cosecθ - sinθ)(secθ - cosθ) = 1/(tanθ + cotθ)

To Find

The proof of the statement

Solution

Here we need to use the following formulas

• Formula 1 = tanx = sinx/cosx
• Formula 2 = cotx = cosx/sinx
• Formula 3 = secx = 1/cosx
• Formula 4 = cosecx = 1/sinx
• Formula 5 = sin²x + cos²x = 1

LHS

cosecθ secθ -cosecθ.cosθ - sinθ.secθ  + sinθcosθ

Using Formula 3 and Formula 4 we get

$= \frac{1}{sin\theta cos\theta} -\frac{cos\theta}{sin\theta} - \frac{sin\theta}{cos\theta} + sin\theta cos\theta$

Taking minus common and adding up the second and third terms we get

$= \frac{1}{sin\theta cos\theta} -[ \frac{cos^2\theta + sin^2\theta}{sin\theta cos\theta}] + sin\theta cos\theta$

Using formula 5 we get

$= \frac{1}{sin\theta cos\theta} - \frac{1}{sin\theta cos\theta} + sin\theta cos\theta$

$= sin\theta cos\theta$

RHS

$\frac{1}{tan\theta + cot\theta }$

Using formula 1 and formula  we get

$\frac{1}{\frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta} }$

Adding up the denominator we get

$= \frac{{cos\theta{sin\theta}}}{{sin^2\theta} + {cos^2\theta} }$

Using formula 5 we get

$= {cos\theta{sin\theta}}} }$

Therefore, LHS = RHS. Hence proved.

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