Question

tanA-cotA/sinA. cosA=sec²A-cosec²A

Answer 1

Question :- Prove that

$\rm \: \dfrac{tanA - cotA}{sinA \: cosA} = {sec}^{2}A - {cosec}^{2}A$

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

Consider LHS

$\rm \: \dfrac{tanA - cotA}{sinA \: cosA}$

can be rewritten as

$\rm \: = \: \dfrac{\dfrac{sinA}{cosA} - \dfrac{cosA}{sinA} }{sinA \: cosA}$

$\rm \: = \: \dfrac{ {sin}^{2} A - {cos}^{2} A}{sinA \: cosA} \times \dfrac{1}{sinA \: cosA}$

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

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

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

$\rm \: = \: {sec}^{2}A - {cosec}^{2}A$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \dfrac{tanA - cotA}{sinA \: cosA} = {sec}^{2}A - {cosec}^{2}A \: }}$

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Identities Used :-

$\boxed{ \rm{ \:tanx = \frac{sinx}{cosx} \: \: }}$

$\boxed{ \rm{ \:cotx = \frac{cosx}{sinx} \: \: }}$

$\boxed{ \rm{ \:cosecx = \frac{1}{sinx} \: \: }}$

$\boxed{ \rm{ \:secx = \frac{1}{cosx} \: \: }}$

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Additional Information :-

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

Answer 2

$\large{\underline{\underline{\pmb{\sf{\; Question\; Asked\;:}}}}}$

Prove that:

$\qquad\qquad\bullet\;\sf{\dfrac{tan\; A- cot\; A}{sin\; A. cos\; A}={sec}^{2}\; A-{cosec}^{2}\; A}$

$\large{\underline{\underline{\pmb{\sf{\; Understanding\; the\; question\;:}}}}}$

Here, concept of Trigonometry has been used. We are given with an identity and we have to prove that it's LHS is equal to RHS. So, we'll be using some basic trignometric identities to prove it. We can simplify the LHS and bring it in the form of RHS.

$\large{\underline{\underline{\pmb{\sf{\; Identities\; we'll\; be\; using\; here\;:}}}}}$

$\star\; \boxed{\sf{tan\; A=\dfrac{sin\; A}{cos\; A}}}$

$\star\; \boxed{\sf{cot\; A=\dfrac{cos\; A}{sin\; A}}}$

$\star\; \boxed{\sf{sec\; A=\dfrac{1}{cos\; A}}}$

$\star\; \boxed{\sf{cosec\; A=\dfrac{1}{sin\; A}}}$

$\begin{gathered}\rule{230px}{.2ex}\\\end{gathered}$

$\large{\underline{\underline{\pmb{\sf{\; Solution\;:}}}}}$

Let's take tan A - cot A/sin A . cos A as LHS here.

$\longrightarrow\quad\sf{\dfrac{tan\; A-cot\; A}{sin\; A. cos\; A}=LHS}$

We know,

$\qquad\leadsto\quad\sf{tan\; A=\dfrac{sin\; A}{cos\; A}}$

$\qquad\leadsto\quad\sf{cot\; A=\dfrac{cos\; A}{sin\; A}}$

Using the above identities in the equation we get -

$\longrightarrow\quad\sf{LHS=\dfrac{\dfrac{sin\; A}{cos\; A}-\dfrac{cos\; A}{sin\; A}}{sin\; A. cos\; A}}$

Now, this can be rewritten as -

$\longrightarrow\quad\sf{LHS=\dfrac{{sin}^{2}\; A-{cos}^{2}\; A}{{sin}^{2}\; A. {cos}^{2}\; A}}$

Splitting this we get -

$\longrightarrow\quad\sf{LHS=\dfrac{{sin}^{2}\; A}{{sin}^{2}\; A. {cos}^{2}\; A}-\dfrac{{cos}^{2}\; A}{{sin}^{2}\; A. {cos}^{2}\; A}}$

Cancelling the term in denominator and numerator.

$\longrightarrow\quad\sf{LHS=\dfrac{1}{{cos}^{2}\; A} -\dfrac{1}{{sin}^{2}\; A}}$

We know -

$\qquad\leadsto\quad\sf{sec\; A=\dfrac{1}{cos\; A}}$

$\qquad\leadsto\quad\sf{cosec\; A=\dfrac{1}{sin\; A}}$

Using this identities -

$\longrightarrow\quad\sf{LHS={sec}^{2}\; A-{cosec}^{2}\; A}$

That is RHS. So, LHS = RHS; i.e., $\\\longrightarrow\quad\underline{\boxed{\sf{\dfrac{tan\; A- cot\; A}{sin\; A. cos\; A}={sec}^{2}\; A-{cosec}^{2}\; A}}} \; \bigstar$

ㅤ

ㅤHence proved!