Question

without using trigonometric tables evaluate the following cosec 39/sec 51 +2/root3×tan17×tan38tan60tan52tan73-3 (sin square31 +sin square 59)

Answer 1

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Answer 2

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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