Math, asked by rodajatt001, 1 year ago

If x cot ^2 45 - sec^2 60+ sin^2 30 =1/8 ,then find the value of x

Answers

Answered by Nani11111111
18
Hiii frnd ur answer is
cos45=1,sec60=2,sin30=1/2
x(1)-2+1/4=1/8
x-2=1/8-2/8
x-2=-1/8
x=2-1/8
x=15/8
hope u understand frnd
Answered by payalchatterje
0

Answer:

Required value of x is 3 \frac{7}{8}

Step-by-step explanation:

Given,

x {cot}^{2}  {45}^{o}  -  {sec}^{2}  {60}^{o}  +  {sin}^{2}  {30}^{o}  =  \frac{1}{8}

We know,

cot {(45}^{o} ) = 1 \\ sec( {60}^{o} ) =  2  \\  \sin( {30}^{o} )  =  \frac{1}{2}

So,

x   \times {1}^{2}  -  {2}^{2}  +  { (\frac{1}{2}) }^{2}  =  \frac{1}{8}  \\ x - 4 +  \frac{1}{4}  =  \frac{1}{8}  \\ x =  \frac{1}{8}  + 4 -  \frac{1}{4}  \\ x =  \frac{1 + 32 - 2}{8}  \\ x =  \frac{31}{8}  \\ x = 3 \frac{7}{8}

Some important formulas of Trigonometry,

sin(x)  =  \cos(\frac{\pi}{2}  - x)  \\  \tan(x)  =  \cot(\frac{\pi}{2}  - x)  \\  \sec(x)  =  \csc(\frac{\pi}{2}  - x)  \\ \cos(x)  =  \sin(\frac{\pi}{2}  - x)  \\ \cot(x)  =  \tan(\frac{\pi}{2}  - x)  \\ \csc(x)  =  \sec(\frac{\pi}{2}  - x)

know more about Trigonometry,

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