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Find the Value of X In the Following : Square Root Tan2x=Cos60∘+Sin45∘Cos45∘ - Mathematics

Find the value of x in the following :

`sqrt3 tan 2x = cos 60^@ + sin45^@ cos 45^@`

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Solution

We have

`sqrt3 tan 2x = cos 60^@ + sin45^@ cos 45^@`  .......(1)

Now we know that

`sin 45^@ = cos 45^@ = 1/sqrt2 and cos 60^@ = 1/2`

Now by substituting above values in equation (1), we get,

`sqrt3 tan 2x = cos 60^@ + sin 45^@ cos 45^@`

`sqrt3 tan 2x = 1/2 + 1/sqrt2 xx 1/sqrt2`

`= 1/2 + 1/(sqrt2 xx sqrt2)`

`=1/2 + 1/2`

`= (1 + 1)/2`

`= 2/2`

= 1

Therefore,

`sqrt3 tan 2x = 1`

`=> tan 2x = 1/sqrt3`  .....(2)

Since

`tan 30^@ = 1/sqrt3` .....(3)

Therefore by comparing equation (2) and (3)

We get

`2x = 30^@`

`x = 30^@/2`

`=> x = 15^@`

x = 15

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 10 Maths
Chapter 10 Trigonometric Ratios
Exercise 10.2 | Q 24 | Page 42
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