if sec theata +tan theata=p show that p^2-1/o^2+1 = sin theta
Answers
Answer:
We have ,
sec\theta + tan\theta = psecθ+tanθ=p ----(1)
Now,
sec^{2}\theta-tan^{2}\theta = 1sec
2
θ−tan
2
θ=1
\implies (sec\theta + tan\theta)(sec\theta - tan\theta)=1⟹(secθ+tanθ)(secθ−tanθ)=1
\implies p\times (sec\theta - tan\theta)=1⟹p×(secθ−tanθ)=1
\implies sec\theta + tan\theta = \frac{1}{p}⟹secθ+tanθ=
p
1
--(2)
Adding and subtracting (1) and (2) , we get
2sec\theta = \frac{(p^{2}+1)}{p}2secθ=
p
(p
2
+1)
---(3)
2tan\theta = \frac{(p^{2}-1)}{p}2tanθ=
p
(p
2
−1)
---(4)
on dividing equation (4) by (3), we get
\frac{2tan\theta}{2sec\theta}=\frac{\frac{(p^{2}-1)}{p}}{\frac{(p^{2}+1)}{p}}
2secθ
2tanθ
=
p
(p
2
+1)
p
(p
2
−1)
\implies \frac{\frac{sin\theta}{cos\theta}}{\frac{1}{cos\theta}}= \frac{\frac{(p^{2}-1)}{p}}{\frac{(p^{2}+1)}{p}}⟹
cosθ
1
cosθ
sinθ
=
p
(p
2
+1)
p
(p
2
−1)
After cancellation, we get
sin\theta = \frac{\frac{(p^{2}-1)}{p}}{\frac{(p^{2}+1)}{p}}sinθ=
p
(p
2
+1)
p
(p
2
−1)
Answer:
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