Question

sin theta / cos theta (1- cot theta) + cos theta / sin theta ( 1- tan theta ) = 1 + sec theta * cosec theta​

Answer 1

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Prove that

$\sf \: \frac{sin \theta}{cos \theta(1 - cot \theta)} + \frac{cos \theta}{sin \theta(1 - tan \theta)} = 1 + sec \theta \times cosec \theta$

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Taking LHS ,

$\mapsto \rm \: \frac{sin \theta}{cos \theta(1 - cot \theta)} + \frac{cos \theta}{sin \theta(1 - tan \theta)}$

Change cot and tan in sin and cos .

$\mapsto \rm \: \frac{sin \theta}{cos \theta(1 - \frac{cos \theta}{sin \theta} )} + \frac{cos \theta}{sin \theta(1 - \frac{sin \theta}{cos \theta} )}$

Take LCM

$\mapsto \rm \: \frac{sin \theta}{cos \theta( \frac{sin \theta - cos \theta}{sin \theta} )} + \frac{cos \theta}{sin \theta( \frac{cos \theta - sin \theta}{cos \theta} )}$

$\mapsto \rm \: \frac{sin \theta \times sin \theta}{cos \theta(sin \theta -cos \theta)} + \frac{cos \theta \times cos \theta}{sin \theta(cos \theta - sin \theta)}$

$\mapsto \rm \: \frac{ {sin}^{2} \theta}{cos \theta(sin \theta - cos \theta)} - \frac{ {cos}^{2} \theta}{sin \theta(sin \theta - cos \theta)}$

In the above step , we took minus(-) common from denominator of second part .

Now , Take LCM ,

$\mapsto \rm \: \frac{ {sin}^{3} \theta - {cos}^{3} \theta }{sin \theta \times cos \theta(sin \theta - cos \theta)}$

Expand the formula of a³ - b³

$\mapsto \rm \: \frac{ \cancel{(sin \theta - cos \theta)}( {sin}^{2} \theta + {cos}^{2} \theta + sin \theta \: cos \theta )}{sin \theta \: cos \theta \cancel{(sin \theta - cos \theta)}}$

Now , as we know ,

sin² A + cos²A = 1

$\mapsto \rm \: \frac{1 + sin \theta \: cos \theta}{sin \theta \: cos \theta}$

$\mapsto \rm \: \frac{1}{sin \theta \: cos \theta} + \frac{ \cancel{sin \theta \: cos \theta}}{ \cancel{sin \theta \: cos \theta}}$

Now ,

Change $\frac{1}{sinA}$ into cosecA

and

Change $\frac{1}{cosA}$ into secA

$\mapsto \rm \: \frac{1}{sin \theta} \times \frac{1}{cos \theta} + 1$

$\mapsto \rm \: sec \theta \times cosec \theta \: + 1$

As you can see ,

It is equal to RHS .

Hence Proved !

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★ Formula Used :

• $\sf \: cot \: \alpha = \frac{cos \: \alpha }{sin \: \alpha }$
• $\sf \: tan \: \alpha = \frac{sin \: \alpha }{cos \: \alpha }$
• $\sf \: \frac{1}{sin \: \alpha } = coses \: \alpha$
• $\sf \: \frac{1}{cos \: \alpha } = sec \: \alpha$
• a³ - b³ = (a-b)(a²+b²+ab)
• sin² $\alpha$ + cos² $\alpha$ = 1

Answer 2

Answer:

sinθ(1+tanθ)+cosθ(1+cotθ)

=sinθ+ cosθ sin 2θ+cosθ+ sinθcos 2 θ

=sinθ+ sinθcos 2+ cossin 2 θ+cosθA= sinθ

sin

2

θ+cos

2

θ

+

cosθ

sin

2

θ+cos

2

θ

As,sin

2

θ+cos

2

θ=1

A=

sinθ

1

+

cosθ

1

= cosecθ+secθ