Question

Please Help me with this trig! (Trig identities)

Answer 1

Hey there !!

$\begin{lgathered}\bf prove \: that : - \\ \\ \frac{sec \theta + 1}{tan \theta} = \frac{tan \theta}{sec \theta - 1}. \\ \\ \huge solving \: lhs. \\ \\ = \frac{sec \theta + 1}{tan \theta} \\ \\ = \frac{sec \theta + 1}{tan \theta} \times \frac{sec \theta - 1}{sec \theta - 1}. \\ \\ = \frac{ {sec}^{2} \theta - 1}{tan \theta(sec \theta - 1)} . \\ \\ = \frac{ {tan}^{ \cancel2} \theta}{ \cancel{tan \theta}(sec \theta - 1)} . \\ \\ ( \therefore {sec}^{2} \theta - 1 = {tan}^{2} \theta.) \\ \\ \\ \huge \boxed{ \boxed{ \bf = \frac{tan \theta}{sec \theta - 1} .}}\end{lgathered}$
So, 

→ In first blank = sec ∅ - 1 .

→ In second blank = sec²∅ .

→ In third blank = tan²∅.


✔✔ Hence, it is proved ✅✅.

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THANKS


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