Question

Prove cot^2a-cos^2a=cot^2a cos^2a

Answer 1

$\huge\bf\red{\mid{\overline{\underline{Question:}}}\mid}$

Prove that :
${ \cot }^{2}a \: - { \cos}^{2} a = { \cot }^{2} a { \cos }^{2} a$

$\huge\bf\red{\mid{\overline{\underline{Solution :}}}\mid}$

$L.H.S = { \cot }^{2} a - { \cos }^{2}a \\ \\ = {cot}^{2} a - {cot}^{2} a .\:{sin}^{2} a \\ (Since, {cot }^{2}a=\frac {{cos}^{2}a}{{sin}^{2}a} \\ \\ So,\: by \: cross- multiplying, \\ \\ we \: get, {cos}^{2}a= {cot}^{2}a \times {sin}^{2}a)\\ \\ = {cot}^{2} a(1 + {sin}^{2} a) \\ \\ = {cot}^{2} a. {cos}^{2} a = R.H.S$

Hence, Proved.

Answer 2

$\textbf{ \Large { \underline { \underline{QUESTION - }}}}$

$\textbf{cot ^{2} a - cos ^{2} a = \: cot ^{2} a . cos ^{2} a }$

$\textbf{ \Large { \underline { \underline{SOLUTION - }}}}$

$\textbf{ \underline { \underline{we \: have}}}$

$\textbf{cot ^{2} a - cos ^{2} a = \: cot ^{2} a . cos ^{2} a } \\ \\ \textsf{taking \: LHS} \\ \\ \implies{\textbf {cot ^{2} a - cos ^{2} a }} \\ \\ \textsf{as \: we \: know} \\ \\ \rightarrow \: \frac{ \text{cos a}}{\text{sin a}} = \text{cot a} \\ \\ \textsf{by \: cross - multiplying....} \\ \\ \rightarrow \: \text{cos a = cot a . sin a} \\ \\ \textsf{squaring \: both \: side....} \\ \textsf{we \: get...} \\ \\ \text{cos ^{2} a = cot ^{2} a . sin ^{2} a} \\ \\ \boxed{ \textsf{putting \: the \: value \: of \: \text{cos}^{2} a \: in \: \text{cot}^{2} a - \text{cos}^{2} a}} \\ \\ \implies \: \text{cot ^{2} a - cot ^{2} a . sin ^{2} a} \\ \\ \implies \: \text{cot ^{2} a \: (1 - sin ^{2} a)} \\ \\ \textbf{as \: we \: know \: the \: 1st \: trigonometrical identity} \\ \\ \rightarrow \: \text{sin ^{2} a + cos ^{2} a = 1} \\ \rightarrow \text{cos ^{2} a = 1 - sin ^{2} a} \\ \\ \boxed{ \textsf{putting \: this \: value \: of \: cos ^{2} a \: in \: cot ^{2} a \: (1 - sin ^{2} a)}} \\ \\ \text{cot ^{2} a \: (1 - sin ^{2} a)} \\ \\ \implies \text{cot ^{2} a . cos ^{2} a}$

$\boxed{ \text{cot ^{2} a . cos ^{2} a = RHS}}$

$\textbf { \underline { \underline{hence \: proved}}}$