Question

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Prove the given questions in attachment!!​

Answer 1

Answer:

Step-by-step explanation:

Here u go

Sry abt the handwriting

Hope it is helpful

(2 and 3 )

Answer 2

$\huge \bold{ANSWER}$

◼$\huge \boxed{\frac{sin \theta}{1- cot \theta} + \frac{cos\theta}{1- tan \theta} = cos \theta + sin \theta}$

On taking left hand side,

▶$\large{\frac{sin \theta}{1- cot \theta} + \frac{cos\theta}{1- tan \theta}}$

Since,

$\boxed{tan\theta = \frac{sin \theta }{cos \theta}}$

and

$\boxed{cot\theta = \frac{cos \theta }{sin \theta}}$

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

▶$\frac{sin \theta}{ \frac{sin \theta - cos \theta}{sin \theta}}+ \frac{cos \theta}{ \frac{cos \theta - sin \theta}{cos \theta}}$

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

▶$\frac{{sin}^{2} \theta}{cos \theta - sin \theta} - \frac{{cos}^{2} \theta}{cos \theta - sin \theta}$

▶$\frac{{cos}^{2} - {sin}^{2}}{cos \theta - sin \theta}$

▶$\frac{(cos \theta - sin \theta )(cos \theta + sin \theta)}{cos \theta - sin \theta}$

▶$\frac{ \cancel{(cos \theta - sin \theta )}(cos \theta + sin \theta)}{ \cancel{cos \theta - sin \theta}}$

▶$cos \theta + sin \theta$ = R.H.S

Hence Proved!!!!

◼$\huge \boxed{ \frac{sin \theta - 2 {sin}^{3}\theta}{2 {cos}^{3} - cos \theta}}$

On taking sin as common in numerator and cos as common in denominator,

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

Since,

$\boxed{ \frac{sin \theta }{cos \theta}= tan\theta}$

and

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

▶$tan\theta\times\frac{(1-2 \times (1- {cos}^{2} \theta)}{cos \theta(2 {cos}^{2} \theta -1)}$

▶$tan\theta\times\frac{(1-2 + 2 {cos}^{2} \theta)}{(2 {cos}^{2} \theta -1)}$

▶$tan\theta\times \frac{\cancel{(-1+ 2 {cos}^{2} \theta)}}{\cancel {(2 {cos}^{2} \theta -1)}}$

▶$tan \theta$=R.H.S

HENCE PROVED!!

$\huge \boxed{(1+cot \theta - cosec \theta)\times(1-tan \theta + sec \theta) =2}$

On taking left hand side ,

Since,

$\boxed{cot \theta =\frac{cos\theta }{sin \theta}}$

$\boxed{tan\theta =\frac{sin\theta }{cos \theta}}$

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

▶$\frac{sin \theta +cos \theta -1}{sin\theta} \times\frac{cos \theta +sin \theta +1}{cos\theta}$

On multiplying,

▶$\dfrac{sin \theta \times cos \theta+ {sin}^{2} \theta - sin \theta + {cos}^{2} \theta + sin \theta \times cos \theta + cos \theta- cos \theta -sin\theta-1}{sin \theta \times cos \theta}$

Since,

${sin}^{2} + {cos}^{2} =1$

On solving we get,

▶$\frac{2 sin \theta \times cos \theta+ 1- 1}{sin \theta \times cos \theta}$

▶$\frac{2 \times \cancel{sin \theta \times cos \theta}+ 1- 1}{ \cancel{sin \theta \times cos \theta}}$

▶2 =R.H.S

HENCE PROVED!