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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 4...

Evaluate23cos⁡ec258−23cot⁡58.tan⁡32−53tan⁡13tan⁡37tan⁡45tan⁡53tan⁡77\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.

Explanation

Solution

Hint: In these types of questions use the transformation formula and some basic concepts of trigonometry.
23cos⁡ec258−23cot⁡58.tan⁡32−53tan⁡13tan⁡37tan⁡45tan⁡53tan⁡77\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)
23cos⁡ec258−23cot⁡58tan⁡32−53tan⁡13tan⁡37tan⁡45tan⁡(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\tan (90 - 37)\tan (90 - 13)
=23cos⁡ec258−23cot⁡58tan⁡32−53tan⁡13tan⁡37tan⁡45cot⁡37cot⁡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)
=23cos⁡ec258−23cot⁡58tan⁡32−53tan⁡13tan⁡37tan⁡451tan⁡371tan⁡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\dfrac{1}{{\tan 37}}\dfrac{1}{{\tan 13}} (By the formula cot⁡x=1tan⁡x\cot x = \dfrac{1}{{\tan x}})
=23cos⁡ec258−23cot⁡58tan⁡32−53tan⁡45=\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
=23cos⁡ec258−23cot⁡58tan⁡32−53=\dfrac{2}{3}\cos e{c^2}58 - \dfrac{2}{3}\cot 58\tan 32 - \dfrac{5}{3}
=23cos⁡ec258−23cot⁡58tan⁡(90−58)−53=\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)
=23cos⁡ec258−23cot⁡258−53=\dfrac{2}{3}\cos e{c^2}58 - \dfrac{2}{3}{\cot ^2}58 - \dfrac{5}{3}= 23(cos⁡ec258−cot⁡258)−53\dfrac{2}{3}(\cos e{c^2}58 - {\cot ^2}58) - \dfrac{5}{3}
By the trigonometric formula cos⁡ec2x−cot⁡2x=1\cos e{c^2}x - {\cot ^2}x = 1
=23(1)−53=\dfrac{2}{3}(1) - \dfrac{5}{3} {Since(cos⁡ec258−cot⁡258)=1(\cos e{c^2}58 - {\cot ^2}58) = 1}
23−53\dfrac{2}{3} - \dfrac{5}{3}=−1 - 1
So 23cos⁡ec258−23cot⁡58.tan⁡32−53tan⁡13tan⁡37tan⁡45tan⁡53tan⁡77\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 - 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⁡ec2x−cot⁡2x=1\cos e{c^2}x - {\cot ^2}x = 1 to simplify the result as much as possible.