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suynar

· started a discussion

· 1 Months ago

how is sec@-tan@=1/p

Question:

If sec θ + tan θ = P, [P ≠ 0], then tan θ is equal to –

Options:
A) \(\cfrac{1}{2}\left [ \begin{matrix}P+\cfrac{1}{P}\end{matrix} \ \ \right ]\)
B) \(\cfrac{1}{2}\left [ \begin{matrix}P-\cfrac{1}{P}\end{matrix} \ \ \right ]\)
C) \(\cfrac{1}{3}\left [ \begin{matrix}P+\cfrac{1}{P}\end{matrix} \ \ \right ]\)
D) \(\cfrac{1}{3}\left [ \begin{matrix}P-\cfrac{1}{P}\end{matrix} \ \ \right ]\)
Solution:

Ans: (b)


Knowledge Expert

· commented

· 1 Months ago

Dear Student,
According to trignometric identity , Sec^2 x - tan^2 x = 1
( Sec x - tan x) ( sec x - tan x) = 1
( sec x - tan x) * p = 1
( sec x - tan x ) = 1/ p

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