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2/3cosec squre 58°-2/3cot58°tan32°- 5/3tan13°tan37°tan45°tan53°tan77°=? Explain with …

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2/3cosec squre 58°-2/3cot58°tan32°- 5/3tan13°tan37°tan45°tan53°tan77°=? Explain with complete answer please
  • 1 answers

Sia ? 6 years, 6 months ago

According to the question, 
{tex}\frac { 2 } { 3 } \operatorname { cosec } ^ { 2 } 58 ^ { \circ } - \frac { 2 } { 3 } \cot 58 ^ { \circ } \tan 32 ^ { \circ } - \frac { 5 } { 3 }{/tex}{tex} tan13° tan37° tan45° tan53° tan77°{/tex}
{tex}\frac { 2 } { 3 } \operatorname { cosec } ^ { 2 } 58 ^ { \circ } - \frac { 2 } { 3 } \cot 58 ^ { \circ } \tan \left( 90 ^ { \circ } - 58 ^ { \circ } \right) - \frac { 5 } { 3 }{/tex}tan 13° tan 37° tan 45°tan (90° - 37°) tan (90° -13°)
{tex}\frac { 2 } { 3 } \operatorname { cosec } ^ { 2 } 58^\circ - \frac { 2 } { 3 } \cot ^ { 2 } 58^\circ - \frac { 5 } { 3 }{/tex}{tex}tan 13° tan37° tan 45° cot 37° cot 13°{/tex}
{tex}\frac { 2 } { 3 } \left( \operatorname { cosec } ^ { 2 } 58 ^ { \circ } - \cot ^ { 2 } 58 ^ { \circ } \right) - \frac { 5 } { 3 } \tan 13 ^ { \circ } \tan 37 ^ { \circ } \times 1 \times \frac { 1 } { \tan 37 ^ { \circ } } \times \frac { 1 } { \tan 13 ^ { \circ } }{/tex}
{tex}\frac { 2 } { 3 } \times 1 - \frac { 5 } { 3 } = \frac { 2 } { 3 } - \frac { 5 } { 3 } = -1{/tex}

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