Question

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Answer 1

To prove :-

$\rm {tan}^{2}A - {tan}^{2}B =\dfrac{{Cos}^{2} B - {Cos}^{2} A }{{Cos}^{2} A {Cos}^{2}B }$

Identity used:-

• $\bold\green{{tan}^{2}\theta=\dfrac{{Sin}^{2}\theta}{{Cos}^{2}\theta}}$

• $\bold\green{{Sin}^{2}\theta = 1-{Cos}^{2}\theta }$

Answer :-

$\begin{gathered}\rm L.H.S. = {tan}^{2} A - {tan}^{2} B \\\\\rm \: \: \: \: = \dfrac{{Sin}^{2} A }{{Cos}^{2} A} - \dfrac{{Sin}^{2} B }{{Cos}^{2} B } \\\\\rm \: \: \: \: =\dfrac{{Sin}^{2} A {Cos}^{2} B - {Sin}^{2} B {Cos}^{2} A}{{Cos}^{2} A {Cos}^{2} B}\\\\\rm \: \: \: \: =\dfrac{(1-{Cos}^{2} A) {Cos}^{2} B - (1- {Cos}^{2} B) {Cos}^{2} A}{ {Cos}^{2} A {Cos}^{2} B} \\\\\rm \: \: \: \: =\dfrac{{Cos}^{2} B - \cancel{ {Cos}^{2} A {Cos}^{2} B}- {Cos}^{2} A + \cancel{{Cos}^{2} A {Cos}^{2} B} }{ {Cos}^{2} A {Cos}^{2} B}\\\\\rm\green \: \: \: \: = \dfrac{ {Cos}^{2} B - {Cos}^{2} A }{ {Cos}^{2} A {Cos}^{2} B}\\\\\rm\: \: \: \: =R.H.S\: \:[Proved]\end{gathered}$

$\therefore\rm L.H.S = R.H.S [Proved]$

Answer 2

Answer:

To prove :-

\rm {tan}^{2}A - {tan}^{2}B =\dfrac{{Cos}^{2} B - {Cos}^{2} A }{{Cos}^{2} A {Cos}^{2}B }tan2A−tan2B=Cos2ACos2BCos2B−Cos2A

Identity used:-

\bold\green{{tan}^{2}\theta=\dfrac{{Sin}^{2}\theta}{{Cos}^{2}\theta}}tan2θ=Cos2θSin2θ

\bold\green{{Sin}^{2}\theta = 1-{Cos}^{2}\theta }Sin2θ=1−Cos2θ

Answer :-

\begin{gathered}\begin{gathered}\rm L.H.S. = {tan}^{2} A - {tan}^{2} B \\\\\rm \: \: \: \: = \dfrac{{Sin}^{2} A }{{Cos}^{2} A} - \dfrac{{Sin}^{2} B }{{Cos}^{2} B } \\\\\rm \: \: \: \: =\dfrac{{Sin}^{2} A {Cos}^{2} B - {Sin}^{2} B {Cos}^{2} A}{{Cos}^{2} A {Cos}^{2} B}\\\\\rm \: \: \: \: =\dfrac{(1-{Cos}^{2} A) {Cos}^{2} B - (1- {Cos}^{2} B) {Cos}^{2} A}{ {Cos}^{2} A {Cos}^{2} B} \\\\\rm \: \: \: \: =\dfrac{{Cos}^{2} B - \cancel{ {Cos}^{2} A {Cos}^{2} B}- {Cos}^{2} A + \cancel{{Cos}^{2} A {Cos}^{2} B} }{ {Cos}^{2} A {Cos}^{2} B}\\\\\rm\green \: \: \: \: = \dfrac{ {Cos}^{2} B - {Cos}^{2} A }{ {Cos}^{2} A {Cos}^{2} B}\\\\\rm\: \: \: \: =R.H.S\: \:[Proved]\end{gathered}\end{gathered}L.H.S.=tan2A−tan2B=Cos2ASin2A−Cos2BSin2B=Cos2ACos2BSin2ACos2B−Sin2BCos2A=Cos2ACos2B(1−Cos2A)Cos2B−(1−Cos2B)Cos2A=Cos2ACos2BCos2B−Cos2ACos2B−Cos2A+Cos2ACos2B=Cos2ACos2BCos2B−Cos2A=R.H.S[Proved]

\therefore\rm L.H.S = R.H.S [Proved]∴L.H.S=R.H.S[Proved]