Question

(1 + cot theta + tan theta ) (sin theta minus cos theta )divide by sec cube minus cosec cube ​

Answer 1

Answer:

Solution :

From Left hand side

( 1 + cotФ + tanФ ) ( sinФ - cosФ ) = ( 1 +\dfrac{sin\Theta }{cos\Theta }+\dfrac{cos\Theta }{sin\Theta })(1+

cosΘ

sinΘ

+

sinΘ

cosΘ

) ( sinФ - cosФ )

= \dfrac{sin\Theta cos\Theta+sin^{2}\Theta +cos^{2}\Theta }{sin\Theta cos\Theta }

sinΘcosΘ

sinΘcosΘ+sin

2

Θ+cos

2

Θ

( sinФ - cosФ )

= \dfrac{sin\Theta cos\Theta+1}{sin\Theta cos\Theta }

sinΘcosΘ

sinΘcosΘ+1

( sinФ - cosФ )

= sinФ ( \dfrac{sin\Theta cos\Theta+1}{sin\Theta cos\Theta }

sinΘcosΘ

sinΘcosΘ+1

) - cosФ ( \dfrac{sin\Theta cos\Theta+1}{sin\Theta cos\Theta }

sinΘcosΘ

sinΘcosΘ+1

)

= ( \dfrac{1}{cos\Theta }

cosΘ

1

+ sinФ ) - ( \dfrac{1}{sin\Theta }

sinΘ

1

+ cosФ )

= secФ + sinФ - cosecФ - cosФ .......1

Now,

From Right hand side

\dfrac{sec\Theta }{cosec^{2}\Theta }

cosec

2

Θ

secΘ

- \dfrac{cosec\Theta }{sec^{2}\Theta }

sec

2

Θ

cosecΘ

= \dfrac{\dfrac{1}{cos\Theta }}{\dfrac{1}{sin^{2}\Theta }}

sin

2

Θ

1

cosΘ

1

- \dfrac{\dfrac{1}{sin\Theta }}{\dfrac{1}{cos^{2}\Theta }}

cos

2

Θ

1

sinΘ

1

= \dfrac{sin^{2}\Theta }{cos\Theta }

cosΘ

sin

2

Θ

- \dfrac{cos^{2}\Theta }{sin\Theta }

sinΘ

cos

2

Θ

= \dfrac{sin^{3}\Theta -cos^{3}\Theta }{cos\Theta sin\Theta }

cosΘsinΘ

sin

3

Θ−cos

3

Θ

= \dfrac{(sin\Theta -cos\Theta )(sin^{2}\Theta +cos^{2}\Theta +sin\Theta cos\Theta )}{sin\Theta cos\Theta }

sinΘcosΘ

(sinΘ−cosΘ)(sin

2

Θ+cos

2

Θ+sinΘcosΘ)

= \dfrac{(sin\Theta -cos\Theta )(1 +sin\Theta cos\Theta )}{sin\Theta cos\Theta }

sinΘcosΘ

(sinΘ−cosΘ)(1+sinΘcosΘ)

= \dfrac{1}{cos\Theta }

cosΘ

1

+ sinФ - \dfrac{1}{sin\Theta }

sinΘ

1

+ cosФ

= secФ + sinФ - cosecФ - cosФ .........2

So From 1 and 2

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