Question

If cos A = 7/25 (3pi)/2 < A < 2 pi and , then find the value of cot A / 2 .​

Answer 1

• refer attachment for detailed answers

• taking in mind in 4th quadrant sign will be - for cot and tan

Answer 2

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

Given that,

$\rm :\longmapsto\:\dfrac{3\pi}{2} < A < 2\pi$

and

$\rm :\longmapsto\:cosA = \dfrac{7}{25}$

Now, we know that,

$\purple{\rm :\longmapsto\:\boxed{\tt{ cos2x = \frac{1 - {tan}^{2}x}{1 + {tan}^{2}x}}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{1 - {tan}^{2} \dfrac{A}{2} }{1 + {tan}^{2} \dfrac{A}{2}} = \dfrac{7}{25}$

$\rm :\longmapsto\:7 + 7{tan}^{2} \dfrac{A}{2} = 25 - 25{tan}^{2} \dfrac{A}{2}$

$\rm :\longmapsto\:7{tan}^{2} \dfrac{A}{2} + 25{tan}^{2} \dfrac{A}{2} = 25 - 7$

$\rm :\longmapsto\:32{tan}^{2} \dfrac{A}{2} = 18$

$\rm :\longmapsto\:16{tan}^{2} \dfrac{A}{2} = 9$

$\rm :\longmapsto\:{tan}^{2} \dfrac{A}{2} = \dfrac{9}{16}$

$\bf\implies \:{tan}\dfrac{A}{2} = \: \pm \: \dfrac{3}{4}$

Now, As

$\rm :\longmapsto\:\dfrac{3\pi}{2} < A < 2\pi$

$\rm\implies \:\dfrac{3\pi}{4} < \dfrac{A}{2} < \pi$

$\rm\implies \: \dfrac{A}{2} \: \in \: {2}^{nd} \: quadrant$

$\rm\implies \: tan\dfrac{A}{2} < 0$

$\bf\implies \:tan \dfrac{A}{2} \: = \: - \: \dfrac{3}{4}$

$\bf\implies \:cot \dfrac{A}{2} \: = \: - \: \dfrac{4}{3}$

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MORE TO KNOW

$\boxed{\tt{ sin2x = 2sinxcosx = \frac{2tanx}{1 + {tan}^{2} x}}}$

$\boxed{\tt{ cos2x = {2cos}^{2}x - 1 = 1 - {2sin}^{2}x}}$

$\boxed{\tt{ cos2x = {cos}^{2}x - {sin}^{2}x = \frac{1 - {tan}^{2}x}{1 + {tan}^{2}x}}}$

$\boxed{\tt{ tan2x = \frac{2tanx}{1 - {tan}^{2}x }}}$

$\boxed{\tt{ 1 - cos2x = {2sin}^{2}x}}$

$\boxed{\tt{ 1 + cos2x = {2cos}^{2}x}}$