Question

without using trigonometric tables, evaluate the following :-) [tan 20°/ cosec 70°]^2 + [ cot 20°/ sec 70°]^2 +2 tan15° tan45° tan75°

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

$\left(\frac{tan20}{cosec70}\right)^{2}+\left(\frac{cot20}{sec70}\right)^{2}+2tan15tan45tan75=3$

Step-by-step explanation:

$Value \:of \: \left(\frac{tan20}{cosec70}\right)^{2}+\left(\frac{cot20}{sec70}\right)^{2}+2tan15tan45tan75$

$= \left(\frac{tan20}{cosec(90-20)}\right)^{2}+\left(\frac{cot20}{sec(90-20)}\right)^{2}+2tan(90-75)tan75tan45$

$= \left(\frac{tan20}{sec20}\right)^{2}+\left(\frac{cot20}{cosec20}\right)^{2}+2(cot75tan75)\times 1$

$= \left(\frac{\frac{sin20}{cos20}}{\frac{1}{cos20}}\right)^{2}+\left(\frac{\frac{cos20}{sin20}}{\frac{1}{sin20}}\right)^{2}+2\times 1$

$=sin^{2}20 + cos^{2}20+2$

$= 1+2\\=3$

Therefore,

$\left(\frac{tan20}{cosec70}\right)^{2}+\left(\frac{cot20}{sec70}\right)^{2}+2tan15tan45tan75=3$