Question

Prove that Sin^2 pie/6+ cos^2pie/3-tan^2pie/4=-1/2

Answer 1

Answer:

Sin 2π\6+Cos 2π\3-Tan 2π\4

Step-by-step explanation:

sin 60+cos120+tan 90

√3/2-1/2+1/0

Answer 2

$\large{\overline{\underline{ \sf \red{\bigstar \: Question: -}}}} </p><p></p><p>$

Prove that :-

• $\large\sf { \sin }^{2} \dfrac{\pi}{6} + { \cos}^{2} \dfrac{\pi}{3} - { \tan }^{2} \dfrac{ \pi}{4} =\dfrac{-1}{2}$

$\large{\overline{\underline{ \sf \red{\bigstar \: Solution : -}}}} </p><p></p><p>$

Taking LHS :-

• $\large= { \sin}^{2} \dfrac{ \pi}{6} + { \cos }^{2} \dfrac{ \pi}{6} - { \tan }^{2} \dfrac{ \pi}{4} </li></ul><p> \:$
  • $\large= {\bigg( \dfrac{1}{2} \bigg)}^{2} + {\bigg( \dfrac{1}{2} \bigg)}^{2} - ( {1})^{2}$
  • $\large= \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{1}$
  • $\large= \dfrac{2 - 4}{4}$
  • $\large = \dfrac{ \cancel { - 2}}{ \cancel4}$
  • $\large = \dfrac{ - 1}{2}$
  • = RHS

  LHS = RHS

  Hence proved !!