Question

(1+cot theta - cosec theta) (1+tan theta + sec theta) prove it . please tell me it's urgent​

Answer 1

To Prove :

(1+cot theta - cosec theta) (1+tan theta + sec theta) = 2

Proof :

The LHS is given as :

(1+cot theta - cosec theta) (1+tan theta + sec theta)

We need to simplify this first .

To do this , first , we need to bring all the terms to the form of sin theta and cos theta.

(1+cot theta - cosec theta) (1+tan theta + sec theta)

=> [ 1 + {cos theta }/{ sin theta } - { 1}/{ sin theta } ][ 1 + {sin theta}/{cos theta} + {1}/{ cos theta } ]

Now , carefully notice the above product .

For (1+cot theta - cosec theta) part , we can take LCM with the denominator as sin theta .

For (1+tan theta + sec theta) part , we can take LCM with the denominator as cos theta .

So ,

=>{ [ sin theta + cos theta - 1 ] [ cos theta + sin theta + 1 ] } / { sin theta cos theta }

Let us assume that sin theta + cos theta = a .

=> { [ a - 1 ][ a + 1 ] } / { sin theta cos theta } ]

=> [ a² - 1 ] / [ sin theta cos theta ]

=> [ ( sin theta + cos theta )² - 1 ] / [ sin theta cos theta ]

=> [ sin ² theta + cos² theta + 2sin theta cos theta - 1 ] / [ sin theta cos theta ]

Now, we know that sin² theta + cos² theta is 1 . So , sin² theta + cos² theta and -1 cancel leaving with :

=> [ 2 sin theta cos theta ] / [ sin theta cos theta ]

=> 2 .

Hence Proved

__________________________

Answer 2

Step-by-step explanation:

$\large{(1 - cot - cosec)(1 + tan + sec)}$

$\large(1 - \frac{cos}{sin} - \frac{1}{sin} ) \: (1 + \frac{sin}{cos} + \frac{1}{cos} )$

$\huge \sf \color{cyan} \mathbb{TAKING \: \: LCM}$

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

$\large\frac{sincos + {sin}^{2} + sin - {cos}^{2} - sincos - cos - cos - sin - 1}{sin \: cos}$

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

hope it helps..