Sec39/cosec51 -2/√(3)[tan17 tan38 tan60 tan52 tan73 …
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Posted by Sonu Sonu 4 years, 10 months ago
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Sia ? 4 years, 10 months ago
Given, {tex}\frac { \sec 39 ^ { \circ } } { \operatorname { cosec } 51 ^ { \circ } } + \frac { 2 } { \sqrt { 3 } }{/tex}{tex}tan17° tan38° tan60°tan52°tan73° -3(sin^231° + sin^259°){/tex}
= {tex}\frac { \sec 39 ^ { \circ } } { \operatorname { cosec } \left( 90 ^ { \circ } - 39 ^ { \circ } \right) } + \frac { 2 } { \sqrt { 3 } }{/tex}tan17°tan 38° tan 60°tan (90°-38°) tan (90°-17°) -3(sin231° + sin2(90° - 31°))
{tex}= \frac { \sec 39 ^ { \circ } } { \sec 39 ^ { \circ } } + \frac { 2 } { \sqrt { 3 } } {/tex}tan 17° tan 38° {tex}\times{/tex} {tex}\sqrt{3}{/tex} {tex}\times{/tex} cot 38° {tex}\times{/tex} cot 17° -3(sin231° + cos231°)
= 1 + {tex}\frac { 2 } { \sqrt { 3 } } \times \sqrt { 3 }\times1\times1{/tex} - 3 [{tex}\because tan\theta . cot \theta = 1{/tex}]
{tex}= 1 + 2 - 3\\ = 0{/tex}
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