If secA + tanA= p, show that
secA- tanA =1/p
Hence, find the value of cosA and sinA
please give me the answer of this question soon
Answers
Step-by-step explanation:
Given :-
Sec A + Tan A= p
To find :-
I) Show that Sec A - Tan A = 1/p .
II) Find the values of Cos A and Sin A ?
Solution :-
Given that
Sec A + Tan A= p ------------(1)
We know that
Sec² A - Tan² A = 1
It can be written as
=> ( Sec A + Tan A) (Sec A - Tan A) = 1
Since , (a+b)(a-b) = a²-b²
=> (p) (Sec A - Tan A) = 1 (from (1))
=> Sec A - Tan A = 1/p ---------(2)
Therefore, Sec A - Tan A = 1/p
On adding (1)&(2) then
Sec A + Tan A= p
Sec A - Tan A = 1/p
(+)
________________
2 Sec A + 0 = p+(1/p)
________________
=> 2 Sec A = p + (1/p)
=> 2 Sec A = (p²+1)/p
=> Sec A = (p²+1)/2p
=> 1/Cos A = (p²+1)/2p
=> Cos A = 2p/(p²+1) ---------(3)
On squaring both sides then
=> (Cos A)² = [2p /(p²+1)]²
=> Cos² A = (2p)²/(p²+1)²
=> Cos² A = 4p²/(p²+1)²
On Subtracting above equation from 1 then
=> 1 - Cos² A = 1 - [ 4p²/(p²+1)²]
=> 1 - Cos² A = [(p²+1)²-4p²]/(p²+1)²
=> Sin² A = (p²-1)²/(p²+1)²
Since , (a-b)² = (a+b)²-4ab
Where, a = p² and b = 1
=> Sin² A = [(p²-1)/(p²+1)]²
=> Sin A = (p²-1)/(p²+1)
Therefore, Sin A = (p²-1)/(p²+1)
Answer :-
I) Sec A - Tan A = 1/p
II) Cos A = 2p/(p²+1)
III) Sin A = (p²-1)/(p²+1)
Used Trigonometric Identities :-
→ Sec² A - Tan² A = 1
→ Sin² A + Cos² A = 1
Used Algebraic Identities :-
→ (a-b)² = (a+b)²-4ab
→ (a+b)(a-b) = a²-b²