Question

Help me pls..


evaluate cosec39°/sec51°+2/root3.tan17°.tan52°.tan73-3(sin^2 31°+sin^2 59°
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Answer 1

Answer:

MysticdMaths AryaBhatta

Answer:

\frac{cosec39}{sec51}\\+\frac{2}{\sqrt{3}} tan17 tan38 tan60 tan52 tan73 \\- 3(sin^{2}31+sin^{2} 59)=0

Step-by-step explanation:

\frac{cosec39}{sec51}\\+\frac{2}{\sqrt{3}} tan17 tan38 tan60 tan52 tan73\\ - 3(sin^{2}31+sin^{2} 59)

=\frac{cosec(90-51)}{sec51}\\+\frac{2}{\sqrt{3}} (tan17 tan73) tan60 (tan38 tan52) \\- 3[sin^{2}(90-59)+sin^{2} 59]

=\frac{sec51}{sec51}\\+\frac{2}{\sqrt{3}} tan17 tan(90-17) \times \sqrt{3}\times tan38 tan(90-38)\\ - 3[cos^{2}59+sin^{2} 59]

=1+2\times (tan17 cot17) \times (tan38 cot38) - 3\times 1

=1+2\times 1\times 1-3

= 1+2-3

=0

Therefore,

\frac{cosec39}{sec51}\\+\frac{2}{\sqrt{3}} tan17 tan38 tan60 tan52 tan73 \\- 3(sin^{2}31+sin^{2} 59) = 0

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