Question

Evaluate\[\dfrac{2}{3}\cos e{c^2}58 - \dfrac{2}{3}\cot 58.\tan 32 - \dfrac{5}{3}\tan 13\tan 37\tan 45\tan 53\tan 77\]. All angles in degrees.

Answer 1

Hint: In these types of questions use the transformation formula and some basic concepts of trigonometry.
\[\dfrac{2}{3}\cos e{c^2}58 - \dfrac{2}{3}\cot 58.\tan 32 - \dfrac{5}{3}\tan 13\tan 37\tan 45\tan 53\tan 77\]

Complete step-by-step answer:
Since we know by the trigonometric formula tan (90-x) =cotx (where x is an angle)
So tan77, tan53 can be written as tan (90-13), tan (90-37)
$\dfrac{2}{3}\cos e{c^2}58 - \dfrac{2}{3}\cot 58\tan 32 - \dfrac{5}{3}\tan 13\tan 37\tan 45\tan (90 - 37)\tan (90 - 13)$
$=\dfrac{2}{3}\cos e{c^2}58 - \dfrac{2}{3}\cot 58\tan 32 - \dfrac{5}{3}\tan 13\tan 37\tan 45\cot 37\cot 13$ (By the formula tan (90-x) =cotx)
$=\dfrac{2}{3}\cos e{c^2}58 - \dfrac{2}{3}\cot 58\tan 32 - \dfrac{5}{3}\tan 13\tan 37\tan 45\dfrac{1}{{\tan 37}}\dfrac{1}{{\tan 13}}$ (By the formula $\cot x = \dfrac{1}{{\tan x}}$)
$=\dfrac{2}{3}\cos e{c^2}58 - \dfrac{2}{3}\cot 58\tan 32 - \dfrac{5}{3}\tan 45$
Now putting the value of tan 45 i.e. tan 45= 1
\[=\dfrac{2}{3}\cos e{c^2}58 - \dfrac{2}{3}\cot 58\tan 32 - \dfrac{5}{3}\]
$=\dfrac{2}{3}\cos e{c^2}58 - \dfrac{2}{3}\cot 58\tan (90 - 58) - \dfrac{5}{3}$ (By the formula tan (90-x) =cotx)
$=\dfrac{2}{3}\cos e{c^2}58 - \dfrac{2}{3}{\cot ^2}58 - \dfrac{5}{3}$= $\dfrac{2}{3}(\cos e{c^2}58 - {\cot ^2}58) - \dfrac{5}{3}$
By the trigonometric formula $\cos e{c^2}x - {\cot ^2}x = 1$
$=\dfrac{2}{3}(1) - \dfrac{5}{3}$ {Since$(\cos e{c^2}58 - {\cot ^2}58) = 1$}
$\dfrac{2}{3} - \dfrac{5}{3}$=$ - 1$
So \[\dfrac{2}{3}\cos e{c^2}58 - \dfrac{2}{3}\cot 58.\tan 32 - \dfrac{5}{3}\tan 13\tan 37\tan 45\tan 53\tan 77\] = $ - 1$

Note: In these types of questions use the transformations formulas to simplify the question use trigonometric values like tan45=1 and some trigonometric identities like $\cos e{c^2}x - {\cot ^2}x = 1$ to simplify the result as much as possible.