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Mathematics

If 43\dfrac{4}{3} (sec2 59° - cot2 31°) - 23\dfrac{2}{3} sin 90° + 3 tan2 56° tan2 34° = x3\dfrac{x}{3}, then find the value of x.

Trigonometric Identities

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Answer

Solving the L.H.S of above equation using trigonometric identities,

⇒ 43\dfrac{4}{3}[sec2 59° - cot2 (90 - 59)°] - 23\dfrac{2}{3} sin 90° + 3 tan2 56° tan2 (90 - 56)°

= 43\dfrac{4}{3}(sec2 59° - tan2 59°) - 23\dfrac{2}{3} sin 90° + 3 tan2 56° cot2 56°

= 43\dfrac{4}{3} x 1 - 23\dfrac{2}{3} x 1 + 3 x 1

= 43\dfrac{4}{3} - 23\dfrac{2}{3} + 3

= 23\dfrac{2}{3} + 3

= 113\dfrac{11}{3}.

Comparing it with R.H.S i.e.,

113=x3\dfrac{11}{3} = \dfrac{x}{3}

x = 11.

Hence, the value of x = 11.

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