Question

cos 38° cosec52° ÷cot 18°cot35°cot60°cot55°cot72°+sin square 12°+sin square 78°÷sec square 45°cosec square 45°
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Answer 1

Answer:

$\sqrt{3}+\frac{1}{4}$

Step-by-step explanation:

We  know that,

$sin(\theta) =cos(90\degree-\theta)\\ \\ sin^2(\theta) +cos^2(\theta) = 1 \\ \\ cot(\theta) = tan(90\degree-\theta) \\cos(\theta) = sin(90\degree-\theta) \\ \\ sec^2(45\degree)=cosec^2(45\degree) = 2$

Now,

$\frac{cos(38\degree)cosec(52\degree)}{cot(18\degree)*cot(35\degree)cot(60\degree)cot(55\degree)cot(72\degree)}+\frac{sin^2(12\degree)+sin^2(78\degree)}{sec^2(45\degree)+cosec(45\degree)} \\ \\ =>\frac{cos(38\degree)sec(38\degree)}{cot(18\degree)*cot(35\degree)cot(60\degree)tan(35\degree)tan(72\degree)}+\frac{sin^2(12\degree)+cos^2(12\degree)}{2+2} \\ \\ => \frac{1}{cot(60\degree)}+\frac{1}{4} \\ \\ => \sqrt{3} + \frac{1}{4}$

Answer 2

Answer:

here is your required answer,