Math, asked by sentibenimchen, 9 months ago

prove that : sin² pie/6 + cos²pie/3 - tan²pie/4 = - 1/4​

Answers

Answered by nayakjayashree937
4

Answer:

Taking LHS

sin ²pie /6+ cos²pie/3- tan²pie/4

Putting pie = 180'

sin²180/6+cos²180/3- tan²180/6

in²30'+cos²60'- tan²45'

(sin 30')²+(cos 60')²- (tan 45')²

Putting

sin30'=1/2

cos60'=1/2

tan45'=1

=(1/2)²+(1/2)²- (1)²

=1/4+1/4-1

=1+1-6/4

=2-4/4

= -2/4

= - 1/2

RHS

Hence proved.

Answered by MisterIncredible
13

Question : -

Prove that : sin² π/6 + cos² π/3 - tan² π/4 = -1/2

ANSWER

Given : -

sin² π/6 + cos² π/3 - tan² π/4 = -1/4

Required to prove : -

  • LHS = RHS

Concept used : -

The standard values trigonometry angles are ;

 \large{ \bf{\begin{tabular}{|c|c|c|c|c|c|} \cline{1-6}  \theta &  { \tt 0}^{ \circ} &  { \tt 30}^{ \circ} & { \tt 45}^{ \circ} & { \tt 60}^{ \circ} &  { \tt 90}^{ \circ}   \\ \cline{1-6} \sin &0 & $\dfrac{1}{2}$  &$ \dfrac{1}{ \sqrt{2} }  $&$ \dfrac{ \sqrt{3} }{2}  $& 1 \\  \cline{1 - 6} \cos & 1 & $\dfrac{\sqrt{3}}{2}$ & $\dfrac{1}{ \sqrt{2}}$ &$ \dfrac{1}{2}$ & 0 \\  \cline{1 - 6} \tan &$ 0 $ &  $ \dfrac{1}{ \sqrt{3} }   $ & $  1 $ &  \sqrt{3} &   \infty \\  \cline{1 - 6} \end{tabular}}}

Proof : -

sin² π/6 + cos² π/3 - tan² π/4 = -1/2

We need to prove that LHS = RHS .

So,

We know that ;

\tt{ \pi^c = 180^{\circ} }

This implies ;

sin² 180°/6 + cos² 180°/3 - tan² 180°/4 = -1/2

sin² 30° + cos² 60° - tan 45° = -1/2

From the standard values of trigonometric angles we can say that ;

  • sin 30° = 1/2

  • cos 60° = 1/2

  • tan 45° = 1

By substituting these values in the above one .

( 1/2 )² + ( 1/2 )² - (1 )² = -1/2

1/4 + 1/4 - 1 = -1/2

1 + 1/4 - 1 = -1/2

2/4 - 1 = -1/2

2/4 - 4/4 = -1/2

- 2/4 = -1/2

-1/2 = -1/2

LHS = RHS

Hence Proved !

- : Additional Information : -

Some important formulae ;

sin ( A + B ) = sin A cos B + sin B cos A

sin ( A - B ) = sin A cos B - sin B cos A

cos ( A + B ) = cos A cos B - sin A sin B

cos ( A - B ) = cos A cos B + sin A sin B

tan ( A + B ) = tan A + tan B/1 - tan A tan B

tan ( A - B ) = tan A - tan B/1 + tan A tan B

___________________________________

sin @ = opposite side/Hypotenuse

cos @ = Adjacent side/Hypotenuse

tan @ = Opposite side/Adjacent side

cot @ = Adjacent side/Opposite side

sec @ = Hypotenuse/Adjacent side

cosec @ = Hypotenuse/Opposite side

Here, ( @ = theta )

___________________________

Complementary angles

sin @ = cos ( 90° - @ )

cos @ = sin ( 90° - @ )

tan @ = cot ( 90° - @ )

cot @ = tan ( 90° - @ )

sec @ = cosec ( 90° - @ )

cosec @ = sec ( 90° - @ )

Here , ( @ = theta )

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