Question

solve 4 into sec square 59 and minus cos square 31 by 3 into minus 2 by 3 sin 90 + 3 tan square 56 ×tan square 34 equal to X by 3

Answer 1

$\huge\star{\red{\underline{\mathfrak{Answer}}}}$

x=11

☆Step by step explaination:-

In this question we need to use three concepts :

1) trigonometric ratios of some specific angles

2) trigonometric ratios of complementary angles

3)Trigonometric identities

SOLUTION:-

$4( \frac{ { \sec }^{2}59 - { \cot }^{2}31 }{3} ) - \frac{2}{3} \sin(90 ) + 3 { \tan}^{2} 56 \times { \tan}^{2} 34 = \frac{x}{3}$

$4(\frac{{cosec}^{2} (90 - 59) - { \cot }^{2} 31}{3}) - \frac{2}{3} (1) + 3 { \cot}^{2} (90 - 56) \times { \tan}^{2} 34 = \frac{x}{3}$

$4( \frac{{ \cosec }^{2} 31 - { \cot} ^{2} 31}{3}) - \frac{2}{3} + 3( \frac{1}{ \tan {}^{2} 34 } ) \times { \tan}^{2} 34 = \frac{x}{3}$

$4( \frac{1}{3} ) - \frac{2}{3} + 3 = \frac{x}{3}$

$\frac{4}{3} - \frac{2}{3} + 3 = \frac{x}{3}$

$\frac{4 - 2}{3} + 3 = \frac{x}{3}$

$\frac{2}{3} + \frac{3}{1} = \frac{x}{3}$

$\frac{2 + 9}{3} = \frac{x}{3}$

$\frac{11}{3} = \frac{x}{3}$

$33 = 3x$

$\huge\boxed{\texttt{\fcolorbox{black}{yellow}{x=3}}}$

${\pink{\underline{\mathfrak{Important\:formulas\:used:}}}}$

●$\sec(90 - \theta) = \csc( \theta)$

●$sin(90) = 1$

●$\tan(90 - \theta) = \cot( \theta)$

●${ \cot}^{2} \theta + 1 = \csc ^{2} \theta$

●$cot \theta = \frac{1}{tan \theta}$