Question

Good afternoon Friends,

Plz, answer my question


If sec theta + tan theta = m, show that ( m² - 1 )/( m² + 1 ) = sin theta .

Answer 1

Hey there !!


▶ Prove that :-

If sec∅ + tan∅ = m, show that $\frac{( {m}^{2} - 1)}{( {m}^{2} + 1) } .$ = sin²∅ .


▶ Solution :-

Given, sec∅ + tan∅ = m.


➡ Solving RHS,

We have

→ ( m² - 1 ) .

= ( sec∅ + tan∅ )² - 1.

= sec²∅ + tan²∅ + 2sec∅tan∅ - 1 .

= ( sec²∅ - 1 ) + tan²∅ + 2sec∅tan∅ .

= tan²∅ + tan²∅ + 2sec∅tan∅ .

= 2tan²∅ + 2sec∅tan∅ .

= 2tan∅( tan∅ + sec∅ ) ............(1).


→ ( m² + 1 ).

= ( sec∅ + tan∅ )² + 1.

= sec²∅ + tan²∅ + 2sec∅tan∅ + 1 .

= ( tan²∅ + 1 ) + sec²∅ + 2sec∅tan∅ .

= sec²∅ + sec²∅ + 2sec∅tan∅ .

= 2sec²∅ + 2sec∅tan∅ .

= 2sec∅( sec∅ + tan∅ ) ............(2).


▶Now,

= $\frac{( {m}^{2} - 1)}{( {m}^{2} + 1) } .$

$= \frac{ \cancel2tan \theta \cancel{(tan \theta + sec \theta)}}{ \cancel2sec \theta \cancel{(sec \theta + tan \theta )}} .$


$= \frac{tan \theta}{sec \theta} \: .$


$\huge = \frac{ \frac{sin \theta}{ \cancel{cos \theta}} }{ \frac{1}{ \cancel{cos \theta}} } .$



$\huge \boxed{ \boxed{ \bf = sin \theta. }}$


$\huge \bf \underline{ \mathbb{LHS = RHS.}}$



✔✔ Hence, it is proved ✅✅.

____________________________________


THANKS


#BeBrainly.

Answer 2

$\boxed{Answer :}$


$\textbf{Method 1.}$

[Note :- Q = theta]

secQ + tanQ = m

Squaring on both sides,

(secQ + tanQ)² = m²

sec²Q + tan²Q + 2 secQ tanQ = m²


Now,

[(m² - 1) ÷ (m² + 1)]

So,

(m² - 1) = [(sec²Q + tan²Q + 2 secQ tanQ) - 1]

= sec²Q + tan²Q + 2 secQ tanQ - 1

Now, sec²Q - 1 = tan²Q

So,

= tan²Q + tan²Q + 2 secQ tanQ

= 2 tan²Q + 2 secQ tanQ

= 2 tanQ (tanQ + secQ) .......(A)


And

(m² + 1) = [(sec²Q + tan²Q + 2 secQ tanQ) + 1]

= sec²Q + tan²Q + 2 secQ tanQ + 1

= sec²Q + sec²Q - 1 + 2 secQ tanQ + 1

As,

sec²Q - 1 = tan²Q

= 2 sec²Q + 2 secQ tanQ

= 2 secQ (secQ + tanQ) .......(B)

Now;

[(m² - 1) ÷ (m² + 1)]

= [2 tanQ (secQ + tanQ) ÷ 2 secQ (secQ + tanQ)]

= [2 tanQ ÷ 2 secQ]

= (tanQ ÷ secQ)

Now,

tanQ = (sinQ ÷ cosQ)

and sin Q = (1 ÷ cosQ)

So,

[(sinQ ÷ cosQ) ÷ (1 ÷ cosQ)]

= sinQ


$\bf\huge{sinQ = sinQ}$

L.H.S. = R.H.S

Hence, proved
__________________________________

$\textbf{Method 2.}$

$\textbf{In Attachment !!}$

___________________________________