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ASHISH

· started a discussion

· 1 Months ago

Wrong answer (Question: 198223.0 )

let
Sec(square)@= 16x(square)
Tan(square)@= 16/x(square)

1+tan(square)@=Sec(square)@
put a value

1+ 16/x(square) = 16x(square)
1= 16[x(square) - 1/x(square)]
1/16= x(square) - 1/x(square)
Option "A"

Question:
If 4x = sec\(\theta\) and \(\cfrac{4}{x}\) = tan\(\theta\), then 8 \(\left (x^2 - \cfrac {1}{x^2} \right )\) is –
Options:
A) \(\cfrac{1}{16}\)
B) \(\cfrac{1}{8}\)
C) \(\cfrac{1}{2}\)
D) \(\cfrac{1}{4}\)
Solution:

Ans: (c)

you have evaluate the value of 8*(x^2−1/x^2)...............not only (x^2−1/x^2)...............


we have x^2 = sec^2theta/16 and 1/x^2 = tan^2theta/16


so, (x^2−1/x^2) = 1/16


so,   8*(x^2−1/x^2) = 8* 1/16 = 1/2

Knowledge Expert

· commented

· 1 Months ago

Dear Student,

You have to determine the value of 8(x^2-1/x^)........not only (x^2-1/x^2).....

You have just calculated the value of

(x^2-1/x^2) = 1/16

Now , you need to calculate

8(x^2-1/x^) = 8* 1/16 = 1/2

Keep learning with us.
All the Best.
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