Question

prove the following
3 and 4​

Answer 1

Answer:

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Answer 2

HEY MATE YOUR ANSWER IS HERE...

QUESTION 3 :-

TAKING LHS

$\frac{1}{tan \: a \: + cot \: a}$

★BY TRIGNOMATERIC RATIOS★

$tan \: a = \frac{sin \: a}{cos \: a } \\ \\ cot \: a = \frac{cos \: a}{sin \: a}$

HENCE

$= \frac{1}{ \frac{sin \: a}{cos \: a} + \frac{cos \: a}{sin \: a} }$

$= \frac{1}{ \frac{ {sin}^{2} a \: + \: {cos}^{2} a}{sin \: a \: cos \: a} }$

★BY TRIGNOMATERIC IDENTITY★

SIN²∅ + COS²∅ = 1

$= \frac{1}{ \frac{1}{sin \: a \: cos \: a} }$

BY SOLVING IT FURTHER

$sin \: a \: cos \: a$

HENCE PROVED

QUESTION 4 :-

TAKING LHS

$tan \: a - cot \: a$

★BY TRIGNOMATERIC RATIOS★

$tan \: a = \frac{sin \: a}{cos \: a } \\ \\ cot \: a = \frac{cos \: a}{sin \: a}$

HENCE ,

$\frac{sin \: a}{ \: cos \: a} - \frac{cos \: a}{sin \: a}$

BY SOLVING UT FURTHER

$\frac{ {sin}^{2} a \: - \: {cos \: }^{2}a }{sin \: a \: cos \: a}$

★NOW BY TRIGNOMATERIC IDENTITY★

SIN²∅ = 1 - COS²∅

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

BY SOLVING IT FURTHER

$= \frac{1 - 2 {cos}^{2} a }{sin \: a \: cos \: a}$

HENCE PROVED

THANKS FOR YOUR QUESTION HOPE THIS HELPS

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