Question

77.
If sec 0 + tan 0 = p, then sin 0 =​

Answer 1

let theta=x

Given secx+tanx=p-------------------[1]

since sec²x-tan²x=1

=>(secx+tanx)(secx-tanx)=1

=>p(secx-tanx)=1

=>secx-tanx=1/p------------------------>[2]

[1]-[2]

=>2tanx=p-1/p

=(p²-1)/p-------------------------->[3]

[1]+[2]

=>2secx=p+1/p

=(p²+1)/p------------------------->[4]

divide [3] with [4]

=> 2tanx (p²-1)/p

------- = ---------

2secx (p²+1)/p

sinx/cosx p²-1

=> -------------- = ---------

1/cosx p²+1

=> sinx=(p²-1)/(p²+1)

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