Prove that tan square A -- …
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Sia ? 6 years, 6 months ago
{tex}L H S = \tan ^ { 2 } A - \tan ^ { 2 } B{/tex}
{tex}= \frac { \sin ^ { 2 } A } { \cos ^ { 2 } A } - \frac { \sin ^ { 2 } B } { \cos ^ { 2 } B }{/tex}
{tex}= \frac { \sin ^ { 2 } A \cos ^ { 2 } B - \sin ^ { 2 } B \cos ^ { 2 } A } { \cos ^ { 2 } A \cos ^ { 2 } B }{/tex}
{tex}= \frac { \sin ^ { 2 } A \left( 1 - \sin ^ { 2 } B \right) - \sin ^ { 2 } B \left( 1 - \sin ^ { 2 } A \right) } { \cos ^ { 2 } A \cos ^ { 2 } B }{/tex}
{tex}= \frac { \sin ^ { 2 } A - \sin ^ { 2 } A \sin ^ { 2 } B - \sin ^ { 2 } B + \sin ^ { 2 } B \sin ^ { 2 } A } { \cos ^ { 2 } A \cos ^ { 2 } B }{/tex}
{tex}= \frac { \sin ^ { 2 } A - \sin ^ { 2 } B } { \cos ^ { 2 } A \cos ^ { 2 } B } = R H S{/tex}
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