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SHEKHAR RAHEJA

· started a discussion

· 1 Months ago

pls explain (Question: 186119.0 )

when sec@+tan@=4
then how is sec@-tan@=1/4

Question:
If sec\(\theta\) + tan\(\theta\) = 4, (\(\theta\neq\) 90º), then the value of cos\(\theta\) is – 
Options:
A) 0
B) \(\cfrac{8}{17}\)
C) \(\cfrac{17}{8}\)
D) \(\cfrac{4}{5}\)
Solution:
Ans: (b) Sec\(\theta\) + tan\(\theta\)  = 4 

sec\(\theta\) - tan\(\theta\)  = \(\cfrac{1}{4}\)

2 sec\(\theta\) = 4 + \(\cfrac{1}{4} = \cfrac{17}{4}\)

sec \(\theta\)= \(\cfrac{17}{8}\)

cos\(\theta\) = \(\cfrac{8}{17}\)

Knowledge Expert

· commented

· 1 Months ago

Dear student
By using the identity
sec^2 theta - tan^2 theta = 1
we will get sec theta - tan theta = 1/4

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