Question

if secπ+ tanπ = r, then (r^2 -1+2r)/(r^2+1) is equal to
please solve this asap​

Answer 1

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Answer 2

Answer:

(r²-1+ 2r) / (r²+1) = sinθ + cosθ

Step-by-step explanation:

given, secθ + tanθ = r

=> r² = ( secθ + tanθ)²

=> r² = sec²θ + tan²θ + 2secθtanθ

=> r² = sec²θ + (sec²θ-1) + 2secθtanθ

=> r² = 2sec²θ -1 + 2secθtanθ

=> r² = 2secθ(secθ+ tanθ) -1

=> r² = 2r . secθ -1

=> secθ = (r²+1)/2r  -------- (1)

similarly,

r² = ( secθ + tanθ)²

=> r² = sec²θ + tan²θ + 2secθtanθ

=> r² = 1 + tan²θ + tan²θ + 2secθtanθ

=> r² = 1 + 2tan²θ + 2secθtanθ

=> r² = 1 + 2tanθ (tanθ+ secθ)

=> r² = 1 + 2tanθ.r

=> tanθ = (r²-1)/2r ----------------- (2)

so,

secθ / tanθ =  [(r²+1)/2r]  /  [(r²-1)/2r]

=>  (1/cosθ) / (sinθ/cosθ) =  [(r²+1)/2r]  /  [(r²-1)/2r]

=>  (1/sinθ) =  (r²+1)  /  (r²-1)

=> sinθ =  (r²-1) / (r²+1) -------------- (3)

from (1)

secθ = (r²+1)/2r

=> (1/cosθ) = (r²+1)/2r

=> cosθ = 2r / (r²+1) ----------------- (4)

(3) + (4) =>

sinθ + cosθ = (r²-1) / (r²+1)  + 2r / (r²+1)

=> sinθ + cosθ = (r²-1+ 2r) / (r²+1)