Question

13. Prove that cosec A - 2 cot 2A cos A = 2 sin A​

Answer 1

Step-by-step explanation:

We have,

$\tt{cosec(A) - 2 \: cot(2A) \: cos(A)}$

$\tt{ = \dfrac{1}{sin(A)} - \dfrac{2 \: cos(2A)}{sin(2A)} \: cos(A)}$

$\tt{ = \dfrac{1}{sin(A)} - \dfrac{2 \: cos(2A) \:cos(A) }{2 \: sin(A) \: cos(A)} }$

$\tt{ = \dfrac{1}{sin(A)} - \dfrac{ cos(2A) }{ sin(A) } }$

$\tt{ = \dfrac{ 1 - cos(2A) }{ sin(A) } }$

$\tt{ = \dfrac{ 2 \: sin^{2} (A) }{ sin(A) } }$

$\tt{ = 2 \: sin(A) }$

Answer 2

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

Consider,

$\rm :\longmapsto\:cosecA - 2 \: cot2A \: sinA$

can be rewritten as

$\rm \:  =  \: \dfrac{1}{sinA} - 2 \times \dfrac{cos2A}{sin2A} \times sinA$

We know,

$\purple{\rm :\longmapsto\: \boxed{\tt{ sin2x = 2 \: sinx \: cosx \: }}}$

So, using this above expression can be further rewritten as

$\rm \:  =  \: \dfrac{1}{sinA} - 2 \times \dfrac{cos2A}{2 \: sinA \: cosA} \times sinA$

$\rm \:  =  \: \dfrac{1}{sinA} - \dfrac{cos2A}{sinA}$

$\rm \:  =  \: \dfrac{1 - cos2A}{sinA}$

We know,

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

So, using this, we get

$\rm \:  =  \: \dfrac{ {2sin}^{2}A }{sinA}$

$\rm \:  =  \: 2 \: sinA$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ cosecA - 2 \: cot2A \: sinA = 2 \: sinA \: }}$

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MORE TO KNOW

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

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

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

$\boxed{\tt{ sin3x = 3sinx - {4sin}^{3}x}}$

$\boxed{\tt{ cos3x = {4cos}^{3}x - 3cosx}}$

$\boxed{\tt{ tan3x = \frac{3tanx - {tan}^{3} x}{1 - {3tan}^{2} x}}}$