Question

if cosA = 3 sinA prove that sec^2 - sin^2 /tan^2 =cosec^2 -cos^2

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Answer 1

GIVEN :–

• Cos(A) = 3 Sin(A)

TO PROVE :–

$\\ \implies \bf \dfrac{ { \sec}^{2}A - { \sin}^{2}A}{ { \tan}^{2} A} = cosec ^{2} (A) - { \cos}^{2} A$

SOLUTION :–

$\\ \implies \bf \cos(A) = 3 \sin(A)$

• We know that –

$\\ \implies \bf \cos^{2} (A) + \sin ^{2} (A) = 1$

$\\ \implies \bf \sqrt{1 - \ { \sin}^{2} (A) }= 3 \sin(A)$

• Square on both sides –

$\\ \implies \bf {1 - \ { \sin}^{2} (A) }= 9 \sin ^{2} (A)$

$\\ \implies \bf 1= 10\sin ^{2} (A)$

$\\ \implies \bf \sin ^{2} (A) = \dfrac{1}{10}$

$\\ \implies \bf \sin(A) = \dfrac{1}{ \sqrt{10} }$

• And –

$\\ \implies \bf \cos(A) = \dfrac{3}{ \sqrt{10} }$

• And –

$\\ \implies \bf \tan(A) = \dfrac{ \sin(A) }{ \cos(A) }= \dfrac{1}{3}$

• Now let's take L.H.S. –

$\\ \: = \: \bf \dfrac{ { \sec}^{2}A - { \sin}^{2}A}{ { \tan}^{2} A}$

• We should write this as –

$\\ \: = \: \bf \dfrac{ \dfrac{1}{{ \cos}^{2}A}- { \sin}^{2}A}{ { \tan}^{2} A}$

• Now put the values –

$\\ \: = \: \bf \dfrac{ \dfrac{1}{ \left(\dfrac{3}{10} \right) }-\left(\dfrac{1}{10} \right)}{\left(\dfrac{1}{9} \right)}$

$\\ \: = \: \bf \dfrac{ \left(\dfrac{10}{3} \right) -\left(\dfrac{1}{10} \right)}{\left(\dfrac{1}{9} \right)}$

$\\ \: = \: \bf \dfrac{ \left(\dfrac{100 - 3}{3 \times 10} \right)}{\left(\dfrac{1}{9} \right)}$

$\\ \: = \: \bf \dfrac{ \left(\dfrac{97}{10} \right)}{\left(\dfrac{1}{3} \right)}$

$\\ \: = \: \bf \dfrac{97}{10}$

• Now Let's take R.H.S. –

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

$\\ \bf = \dfrac{1}{ \left( \dfrac{1}{10} \right)} - \dfrac{3}{10}$

$\\ \bf = 10- \dfrac{3}{10}$

$\\ \bf = \dfrac{100 - 3}{10}$

$\\ \bf = \dfrac{97}{10}$

$\\ \to \bf L.H.S. = R.H.S.$

$\bf \bf Hence\:\: Proved$