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(Cos^2 20°+cos^2 70°/sec^2 50°-cot^2 40°)+2cosec^2 58°-2cot58° …

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(Cos^2 20°+cos^2 70°/sec^2 50°-cot^2 40°)+2cosec^2 58°-2cot58° tan32°
  • 1 answers

Sia ? 6 years, 4 months ago

We have,
{tex}\frac { \cos ^ { 2 } 20 ^ { \circ } + \cos ^ { 2 } 70 ^ { \circ } } { \sec ^ { 2 } 50 ^ { \circ } - \cot ^ { 2 } 40 ^ { \circ } }{/tex} +2 cosec258° - 2cot58° tan 32° - 4 tan13° tan37° tan45° tan53° tan77°
{tex}\frac { \cos ^ { 2 } 20 ^ { \circ } + \cos ^ { 2 } \left( 90 ^ { \circ } - 20 ^ { \circ } \right) } { \sec ^ { 2 } 50 ^ { \circ } - \cot ^ { 2 } \left( 90 ^ { \circ } - 50 ^ { \circ } \right) }{/tex} + 2cosec258° - 2cot58° tan (90° - 58°)- 4tan13° tan37° tan45° tan(90° - 37°) tan(90° -13°)
{tex}\frac { \cos ^ { 2 } 20 ^ { \circ } + \sin ^ { 2 } 20 ^ { \circ } } { \sec ^ { 2 } 50 ^ { \circ } - \tan ^ { 2 } 50 ^ { \circ } }{/tex} + 2cosec258° - 2cot258° - 4 tan13° tan37° tan45° cot37° cot13°
{tex}\frac { 1 } { 1 }{/tex} + 2(cosec258° - cot258°) - 4(tan13° cot13°) (tan37° cot37°) tan45°
= 1 + 2 - 4 {tex}\times{/tex} 1 {tex}\times{/tex} 1 {tex}\times{/tex} 1 = 3 - 4 = -1

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