Question

$\frac{tantheta + sintheta}{tantheta - sintheta} = \frac{sectheta + 1}{sectheta - 1}$
hey mate plz help me to solve this question​

Answer 1

Answer:

$\mathrm{Prove\:}\dfrac{\tan \left(\theta\right)+\sin \left(\theta\right)}{\tan \left(\theta\right)-\sin \left(\theta\right)}=\dfrac{\sec \left(\theta\right)+1}{\sec \left(\theta\right)-1}:\quad \mathrm{True}$

Step-by-step explanation:

$\dfrac{\tan \left(\theta\right)+\sin \left(\theta\right)}{\tan \left(\theta\right)-\sin \left(\theta\right)}=\dfrac{\sec \left(\theta\right)+1}{\sec \left(\theta\right)-1}$

$\gray{\mathrm{Manipulating\:left\:side}}$

$\gray{\dfrac{\tan \left(\theta\right)+\sin \left(\theta\right)}{\tan \left(\theta\right)-\sin \left(\theta\right)}}$

$\black{\mathrm{Express\:with\:sin,\:cos}}$

$=\dfrac{\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}+\sin \left(\theta\right)}{\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}-\sin \left(\theta\right)}$

$\black{\mathrm{Simplify}\:\dfrac{\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}+\sin \left(\theta\right)}{\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}-\sin \left(\theta\right)}:}$

$\dfrac{\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}+\sin \left(\theta\right)}{\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}-\sin \left(\theta\right)}$

$\gray{\mathrm{Join}\:\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}-\sin \left(\theta\right):\quad \dfrac{\sin \left(\theta\right)-\sin \left(\theta\right)\cos \left(\theta\right)}{\cos \left(\theta\right)}}$

$=\dfrac{\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}+\sin \left(\theta\right)}{\dfrac{\sin \left(\theta\right)-\sin \left(\theta\right)\cos \left(\theta\right)}{\cos \left(\theta\right)}}$

$\gray{\mathrm{Join}\:\dfrac{\sin \left(\theta\right)}{\cos \left(\theta\right)}+\sin \left(\theta\right):\quad \dfrac{\sin \left(\theta\right)+\sin \left(\theta\right)\cos \left(\theta\right)}{\cos \left(\theta\right)}}$

$=\dfrac{\dfrac{\sin \left(\theta\right)+\sin \left(\theta\right)\cos \left(\theta\right)}{\cos \left(\theta\right)}}{\dfrac{\sin \left(\theta\right)-\sin \left(\theta\right)\cos \left(\theta\right)}{\cos \left(\theta\right)}}$

$\gray{\mathrm{Divide\:fractions}:\quad \dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a\cdot \:d}{b\cdot \:c}}$

$=\dfrac{\left(\sin \left(\theta\right)+\sin \left(\theta\right)\cos \left(\theta\right)\right)\cos \left(\theta\right)}{\cos \left(\theta\right)\left(\sin \left(\theta\right)-\sin \left(\theta\right)\cos \left(\theta\right)\right)}$

$\gray{\mathrm{Cancel\:the\:common\:factor:}\:\cos \left(\theta\right)}$

$=\dfrac{\sin \left(\theta\right)+\sin \left(\theta\right)\cos \left(\theta\right)}{\sin \left(\theta\right)-\sin \left(\theta\right)\cos \left(\theta\right)}$

$\gray{\mathrm{Factor\:out\:common\:term\:}\sin \left(\theta\right)}$

$=\dfrac{\sin \left(\theta\right)\left(1+\cos \left(\theta\right)\right)}{\sin \left(\theta\right)-\sin \left(\theta\right)\cos \left(\theta\right)}$

$\gray{\mathrm{Factor\:out\:common\:term\:}\sin \left(\theta\right)}$

$=\dfrac{\sin \left(\theta\right)\left(1+\cos \left(\theta\right)\right)}{\sin \left(\theta\right)\left(1-\cos \left(\theta\right)\right)}$

$\gray{\mathrm{Cancel\:the\:common\:factor:}\:\sin \left(\theta\right)}$

$=\dfrac{1+\cos \left(\theta\right)}{1-\cos \left(\theta\right)}$

$\gray{\mathrm{Use\:the\:following\:identity:}\:\cos \left(x\right)=\dfrac{1}{\sec \left(x\right)}}$

$=\dfrac{1+\dfrac{1}{\sec \left(\theta\right)}}{1-\dfrac{1}{\sec \left(\theta\right)}}$

$\black{\mathrm{Simplify}\:\dfrac{1+\dfrac{1}{\sec \left(\theta\right)}}{1-\dfrac{1}{\sec \left(\theta\right)}}:}$

$\dfrac{1+\dfrac{1}{\sec \left(\theta\right)}}{1-\dfrac{1}{\sec \left(\theta\right)}}$

$\gray{\mathrm{Join}\:1-\dfrac{1}{\sec \left(\theta\right)}:\quad \dfrac{\sec \left(\theta\right)-1}{\sec \left(\theta\right)}}$

$=\dfrac{\dfrac{\sec \left(\theta\right)+1}{\sec \left(\theta\right)}}{\dfrac{\sec \left(\theta\right)-1}{\sec \left(\theta\right)}}$

$\gray{\mathrm{Join}\:1+\dfrac{1}{\sec \left(\theta\right)}:\quad \dfrac{\sec \left(\theta\right)+1}{\sec \left(\theta\right)}}$

$=\dfrac{\dfrac{\sec \left(\theta\right)+1}{\sec \left(\theta\right)}}{\dfrac{\sec \left(\theta\right)-1}{\sec \left(\theta\right)}}$

$\gray{\mathrm{Divide\:fractions}:\quad \dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a\cdot \:d}{b\cdot \:c}}$

$=\dfrac{\left(\sec \left(\theta\right)+1\right)\sec \left(\theta\right)}{\sec \left(\theta\right)\left(\sec \left(\theta\right)-1\right)}$

$\gray{\mathrm{Cancel\:the\:common\:factor:}\:\sec \left(\theta\right)}$

$=\dfrac{\sec \left(\theta\right)+1}{\sec \left(\theta\right)-1}$

$\gray{\mathrm{We\:showed\:that\:the\:two\:sides\:could\:take\:the\:same\:form}}$

$\Rightarrow \mathrm{True}$

Answer 2

Answer:

hey mate please refer to the attachment