Math, asked by sushma3194, 7 months ago

evaluate sin
 {sin}^{2}
2π/3 +cos ^2 5π/6 -Tan ^2 3π/4​

Answers

Answered by anindyaadhikari13
4

Answer:-

 \sf { \sin }^{2}  \big( \frac{2\pi}{3}  \big) +  { \cos}^{2} \big(\frac{5\pi}{6}  \big) -  { \tan}^{2} ( \frac{3\pi}{4} )

 \sf { \sin }^{2}  \big(\pi -  \frac{\pi}{3}  \big) +  { \cos}^{2} \big(\pi - \frac{\pi}{6}  \big) -  { \tan}^{2} ( \pi - \frac{\pi}{4} )

 \sf { \sin }^{2}  \big(\frac{\pi}{3}  \big) +  {  \cos}^{2} \big(- \frac{\pi}{6}  \big) -  { \tan}^{2} (\frac{\pi}{4} )

 \sf =  {\big(\frac{ \sqrt{3} }{2}   \big)}^{2}  +  { \big( \frac{ -  \sqrt{3} }{2}  \big)}^{2}  -  {( - 1)}^{2}

 \sf =  \frac{3}{4}  +  \frac{3}{4}  - 1

 \sf =  \frac{6}{4}  - 1

 \sf =  \frac{6 - 4}{4}

 \sf =  \frac{2}{4}

 \sf =  \frac{1}{2}

 \sf = 0.5

Answered by nehashanbhag0729
4

Answer:

hey watch the above pic for ur answer of the question...

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