Question

7Cos70 / 2sin20 + 3/2 cos55 cosec35 / tan5 tan25 tan45 tan65 tan85

Answer 1

$\dfrac{7\cos 70}{2\sin 20} +\dfrac{3}{2} \dfrac{\cos 55 \csc 35}{\tan 5 \tan 25 \tan 45 \tan 65 \tan85}$ = 5

Step-by-step explanation:

We have,

$\dfrac{7\cos 70}{2\sin 20} +\dfrac{3}{2} \dfrac{\cos 55 \csc 35}{\tan 5 \tan 25 \tan 45 \tan 65 \tan85}$

To find, the value of $\dfrac{7\cos 70}{2\sin 20} +\dfrac{3}{2} \dfrac{\cos 55 \csc 35}{\tan 5 \tan 25 \tan 45 \tan 65 \tan85}$ = ?

∴ $\dfrac{7\cos 70}{2\sin 20} +\dfrac{3}{2} \dfrac{\cos 55 \csc 35}{\tan 5 \tan 25 \tan 45 \tan 65 \tan85}$

$=\dfrac{7\cos (90-20)}{2\sin 20} +\dfrac{3}{2} \dfrac{\cos 55 \csc (90-55)}{\tan 5 \tan 25 \tan 45 \tan (90-25) \tan (90-5)}$

Using the trigonometric identity,

$\sin A= \cos (90-A), \cot A= \tan (90-A)$ and $\secA =\csc (90-55)$

$=\dfrac{7\sin 20}{2\sin 20} +\dfrac{3}{2} \dfrac{\cos 55 \sec 55}{\tan 5 \tan 25 (1) \cot 25 \cot 5}$

$=\dfrac{7}{2} +\dfrac{3}{2} \dfrac{\cos 55\dfrac{1}{\cos 55} }{(\tan 5\cot 5)(\tan 25\cot 25 }$

Using the trigonometric identity,

$\sec A =\dfrac{1}{\cos A}$ and $\tan A\cot A=1$

$=\dfrac{7}{2} +\dfrac{3}{2} \dfrac{1}{(1)(1 )}$

$=\dfrac{7}{2} +\dfrac{3}{2}$

$=\dfrac{7+3}{2} =\dfrac{10}{2}$

= 5

∴ $\dfrac{7\cos 70}{2\sin 20} +\dfrac{3}{2} \dfrac{\cos 55 \csc 35}{\tan 5 \tan 25 \tan 45 \tan 65 \tan85}$ = 5