Question

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

Answer 1

GIVEN :

The equation is $cosec\theta(sec\theta-1)-cot\theta(1-cos\theta)=tan\theta-sin\theta$

TO PROVE :

The equality of the given equation.

SOLUTION :

Given that the equation is $cosec\theta(sec\theta-1)-cot\theta(1-cos\theta)=tan\theta-sin\theta$

We have to prove the given equation $cosec\theta(sec\theta-1)-cot\theta(1-cos\theta)=tan\theta-sin\theta$ is true

That is to prove that LHS=RHS

Now taking the LHS,

$cosec\theta(sec\theta-1)-cot\theta(1-cos\theta)$

By using the properties:

i) $cosecx=\frac{1}{sinx}$

ii) $secx=\frac{1}{cosx}$ and

iii) $cotx=\frac{cosx}{sinx}$

$\frac{1}{sin\theta}(\frac{1}{cos\theta}-1)-\frac{cos\theta}{sin\theta}(1-cos\theta)}$

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

$=\frac{1-cos\theta-cos^2\theta+cos^3\theta}{sin\theta cos\theta}$

By using the formula:

$sin^2x=1-cos^2x$

$=\frac{sin^2\theta-cos\theta(1-cos^2\theta)}{sin\theta cos\theta}$

$=\frac{sin^2\theta-cos\theta(sin^2\theta)}{sin\theta cos\theta}$

$=\frac{sin^2\theta(1-cos\theta)}{sin\theta cos\theta}$

$=\frac{sin\theta(1-cos\theta)}{cos\theta}$

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

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

By using the property:

$tanx=\frac{sinx}{cosx}$

$=tan\theta-sin\theta$=RHS

∴ LHS=RHS

∴ $cosec\theta(sec\theta-1)-cot\theta(1-cos\theta)=tan\theta-sin\theta$

Hence proved.

Answer 2

Answer:

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Step-by-step explanation:

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