Question

Prove that
(cos teta- sin teta+1)/(cos teta+sin teta +1)=cosec teta+cot teta​

Answer 1

Answer:

Step-by-step explanation:

Correct question:-

$\dfrac{cos\theta-sin\theta+1}{cos\theta+sin\theta-1}=cosec\theta+cot\theta$

Solution :-

Taking L.H.S :-

$=\dfrac{cos\theta-sin\theta+1}{cos\theta+sin\theta-1}$

Dividing both numerator and denominator with " sinθ" :-

$=\dfrac{\dfrac{cos\theta-sin\theta+1}{sin\theta}}{\dfrac{cos\theta+sin\theta-1}{sin\theta}}$

$=\dfrac{\dfrac{cos\theta}{sin\theta}-\dfrac{sin\theta}{sin\theta}+\dfrac{1}{sin\theta}}{\dfrac{cos\theta}{sin\theta}+\dfrac{sin\theta}{sin\theta}-\dfrac{1}{sin\theta}}$

$=\dfrac{cot\theta-1+csc\theta}{cot\theta+1-csc\theta}$

$=\dfrac{cot\theta+csc\theta-(csc^2\theta-cot^2\theta)}{cot\theta-csc\theta+1}$

$=\dfrac{cot\theta+csc\theta+(cot^2\theta-csc^2\theta)}{cot\theta-csc\theta+1}$

$=\dfrac{cot\theta+csc\theta+(cot\theta+csc\theta)(cot\theta-csc\theta)}{cot\theta-csc\theta+1}$

$=\dfrac{cot\theta+csc\theta(1+cot\theta-csc\theta)}{1+cot\theta-csc\theta}$

$=cot\theta+cosec\theta$

= R.H.S

Know more :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$