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$\large\ \: Question$ ☟


$\ \textless \ br /\ \textgreater \ ( { \tan }^{2} A \: - { \tan}^{2} B) = \large\frac{( { \sin }^{2} A - { \sin }^{2}B) }{ { \cos }^{2}a \: { \cos }^{2}B} = \large\frac{ { \cos }^{2}B \: - \:{ \cos}^{2}A}{ { \cos }^{2}B \: { \cos }^{2}A}$
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Answer 1

LHS=tan2A−tan2B=cos2sin2A−cos2Bsin2B

LHS=tan2A−tan2B=cos2sin2A−cos2Bsin2B=cos2Acos2Bsin2Acos2B−cos2Asin2B=cos2Acos2B(1−cos2A)cos2A(1−cos2B)

LHS=tan2A−tan2B=cos2sin2A−cos2Bsin2B=cos2Acos2Bsin2Acos2B−cos2Asin2B=cos2Acos2B(1−cos2A)cos2A(1−cos2B)=cos2Acos2Bcos2B−cos2Acos2B−cos2A+cos2Acos2B=cos2Acos2Bcos2B−cos2A

LHS=tan2A−tan2B=cos2sin2A−cos2Bsin2B=cos2Acos2Bsin2Acos2B−cos2Asin2B=cos2Acos2B(1−cos2A)cos2A(1−cos2B)=cos2Acos2Bcos2B−cos2Acos2B−cos2A+cos2Acos2B=cos2Acos2Bcos2B−cos2AAlso 

LHS=tan2A−tan2B=cos2sin2A−cos2Bsin2B=cos2Acos2Bsin2Acos2B−cos2Asin2B=cos2Acos2B(1−cos2A)cos2A(1−cos2B)=cos2Acos2Bcos2B−cos2Acos2B−cos2A+cos2Acos2B=cos2Acos2Bcos2B−cos2AAlso cos

Answer 2

Required Answer is in attachment!

Related Trigonometric identities::

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

Trigonometric ratios::

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

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