Question

if cot thitha=15/8,then evaluate (2+2 sin thitha)(1-sin thitha)/(1+cos thitha)(2-2cos thitha)​

Answer 1

Answer:

=

(1+cosθ)(1−cosθ)

(1+sinθ)(1−sinθ)

=

1−cos

2

θ

1−sin

2

θ

=

sin

2

θ

cos

2

θ

=cot

2

θ

Putting cotθ=

8

15

, we get

cot

2

θ=(

8

15

)

2

=

64

225

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\sf \: cot\theta = \dfrac{15}{8}$

Now, Consider

$\sf \: \dfrac{(2 + 2sin\theta )( 1- sin\theta )}{(1 + cos\theta )(2 - 2cos\theta )}$

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

$\sf \: = \: \dfrac{(1 + sin\theta )(1 - sin\theta )}{(1 + cos\theta )(1 - cos\theta )}$

$\sf \: = \: \dfrac{1 - {sin}^{2} \theta }{1 - {cos}^{2} \theta }$

$\sf \: = \: \dfrac{{cos}^{2} \theta }{ {sin}^{2} \theta }$

$\boxed{ \bf{ \because \: {sin}^{2}x + {cos}^{2}x = 1 \: }}$

$\sf \: = \: {cot}^{2}\theta$

$\sf \: = \: {\bigg(\dfrac{15}{8} \bigg) }^{2}$

$\sf \: = \: \dfrac{225}{64}$

Hence,

$\sf \:\sf \: \implies \: \dfrac{(2 + 2sin\theta )( 1- sin\theta )}{(1 + cos\theta )(2 - 2cos\theta )} = \frac{225}{64}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$