If sec there + tan theta= p then find the value of cos ec theta
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Hey !!
Sec¢ + tan¢ = p ------1)
1/sec¢ + tan¢ ×sec¢ -tan¢/sec¢-tan¢
[Multiplying by sec¢ -tan¢ on numerator and denominator]
Sec¢ - tan¢ /sec²¢ - tan²¢ = 1/p
Sec¢ - tan¢ = 1/p ----------2)
Adding both equation. 1 and 2
We get
Sec¢ + tan¢ + sec¢ - tan¢ = p +1 /p
2sec¢ = p² + 1/p
2Sec¢ = p² + 1 /p----------3)
Now substracting the equation 2 from 1
We get
Sec¢ + tan¢ -(sec¢ - tan¢ ) = p - 1/p
2tan¢ = p² - 1 /p -------4)
Now dividing the equation 4 to 3 we get
2tan¢ /2sec¢ = p² -1 /p /p² + 1 /p
Sin¢/cos¢ ×cos¢ = p² - 1 /p² + 1
Sin¢ = p² - 1 /p² + 1
1/sin¢ = p² + 1 /p² - 1
Cosec¢ = p² + 1 /p² - 1 Answer
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Hope it helps you !!!
@Rajukumar111
Sec¢ + tan¢ = p ------1)
1/sec¢ + tan¢ ×sec¢ -tan¢/sec¢-tan¢
[Multiplying by sec¢ -tan¢ on numerator and denominator]
Sec¢ - tan¢ /sec²¢ - tan²¢ = 1/p
Sec¢ - tan¢ = 1/p ----------2)
Adding both equation. 1 and 2
We get
Sec¢ + tan¢ + sec¢ - tan¢ = p +1 /p
2sec¢ = p² + 1/p
2Sec¢ = p² + 1 /p----------3)
Now substracting the equation 2 from 1
We get
Sec¢ + tan¢ -(sec¢ - tan¢ ) = p - 1/p
2tan¢ = p² - 1 /p -------4)
Now dividing the equation 4 to 3 we get
2tan¢ /2sec¢ = p² -1 /p /p² + 1 /p
Sin¢/cos¢ ×cos¢ = p² - 1 /p² + 1
Sin¢ = p² - 1 /p² + 1
1/sin¢ = p² + 1 /p² - 1
Cosec¢ = p² + 1 /p² - 1 Answer
************************
Hope it helps you !!!
@Rajukumar111
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