Question

Sec theta+tan theta-1/tan theta-sec theta+1=cos theta/1-sin theta

Answer 1

Answer:

Given :

To prove:

Proof:

LHS =

We can write, in place of 1 in the numerator,

Thus, LHS =

By using the formula

Taking secθ+tanθ common in the numerator

Where

We proved the left hand side equal to right hand side by using some of the basic trigonometric identities. If you have all of these formula on your tips, you can solve this as well as the similar questions that involve proving the left hand side equals to the right hand side.

Answer 2

Step-by-step explanation:

$\bf L.H.S = \tt \dfrac{sec\: \theta + tan \: \theta - 1}{tan \: \theta - sec \: \theta + 1}$

$: \implies \tt \dfrac{\frac{1}{cos \: \theta} + \frac{sin \: \theta}{cos \: \theta} - 1}{ \frac{sin \: \theta}{cos \: \theta} - \frac{1}{cos \: \theta} + 1 } \: = \dfrac{1 + sin \: \theta - cos \: \theta}{sin \: \theta + cos \: \theta}$

$: \implies \tt\dfrac{ sin \: \theta - (cos \: \theta - 1)}{sin \: \theta + (cos \: \theta - 1)} \: \times \: \dfrac{ sin \: \theta - (cos \: \theta - 1)}{sin \: \theta - (cos \: \theta - 1)}$

$: \implies \tt\dfrac{ sin^{2} \: \theta + cos^{2} \: \theta + 1 - 2 \: cos \: \theta - 2 \: sin \: \theta \: (cos \: \theta - 1)}{sin^{2} \: \theta - (cos \: \theta - 1)^{2} }$

$: \implies \tt\dfrac{1 + 1 - 2 \: cos \: \theta - 2 \: sin \: \theta \: cos \: \theta + 2 \: sin \: \theta}{sin^{2} \: \theta + cos^{2} \: \theta - 1 + 2 \: cos \: \theta }$

$: \implies \tt\dfrac{2 - 2 \: cos \: \theta - 2 \: sin \: \theta \: cos \: \theta + 2 \: sin \: \theta}{sin^{2} \: \theta + cos^{2} \: \theta - sin^{2} \: \theta - cos^{2} \: \theta + 2 \: cos \: \theta }$

$: \implies \tt\dfrac{2 (1 - \: cos \: \theta )- 2 \: sin \: \theta (1 - \: cos \: \theta)}{ 2 \: cos \: \theta - 2 \: cos^{2} \: \theta}$

$: \implies \tt\dfrac{(2 + 2 \: sin \: \theta) \: \cancel{(1 - cos\: \theta)}}{2 \: cos \: \theta \: \cancel{(1 - cos \: \theta)}} \: = \: \dfrac{1 + sin \: \theta}{cos \: \theta}$

$: \implies\tt\dfrac{1 + sin \: \theta}{cos \: \theta} \: \times \: \dfrac{1 - sin \: \theta}{1 - sin \: \theta}$

$: \implies\tt\dfrac{1 + sin^{2} \: \theta}{cos \: (1 - sin \: \theta)}$

$: \implies\tt\dfrac{cos^{2} \: \theta}{cos \: \theta (1 - sin \: \theta)}$

$: \implies\tt\dfrac{cos \: \theta}{1 - sin \: \theta} \: = \: \bf{ R.H.S}$

$\huge\bigstar \:\underline{\red{\sf Hence, Proved}} \: \bigstar$