Question

If sin thita = a-1 /a+1 , find tan thita
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Answer 1

Answer:

Your answer attached in the photo

Answer 2

The value of $Tan\theta$  is $\frac{a-1}{2\sqrt{a} }$.

Step-by-step explanation:

We are given,

$Sin\theta=\frac{a-1}{a+1}$

and we have to find the value of  $Tan\theta$ .

• Formula used,

we use Pythagoras theorem for solving the given question,

1. $H^{2} =P^{2} +B^{2}$
2. $B=\sqrt{H^{2} -P^{2} }$
3. $(a+1)^{2} =a^{2} +b^{2} +2ab$
4. $(a-1)^{2} =a^{2} +b^{2} -2ab$

$Sin\theta=\frac{P}{H}$    ,    $Cos\theta=\frac{B}{H}$    and     $Tan\theta=\frac{P}{B}$.

H is for Hypotenuse, P is for perpendicular and B is for Base.

• Calculation for $Tan\theta$,

we have,

$Sin\theta=\frac{a-1}{a+1}$  

and $Sin\theta$  is also equal to  $Sin\theta=\frac{P}{H}$ ,

by comparing both $Sin\theta$,  we get,

$\frac{P}{B} =\frac{a-1}{a+1}$

hence we get perpendicular(P) and Base(B),

$P=a-1$   and  $H=a+1$

by using Pythagoras theorem we can calculate the base(B),

$B=\sqrt{H^{2} -P^{2} }$

$B=\sqrt{(a+1)^{2} -(a-1)^{2} }$

by using formula (3) and (4) we get,

$B=\sqrt{[a^{2} +(1)^{2} +2a(1)]-[a^{2} +(1)^{2} -2a(1)]}$

$B=\sqrt{a^{2} +1 +2a-a^{2} -1 +2a}$

$B=\sqrt{4a}$

$B=2\sqrt{a}$

now we have the value of base(B) is $2\sqrt{a}$ .

Calculating $tan\theta$

$Tan\theta=\frac{P}{B}$

we have $P=a-1$  and $B=2\sqrt{a}$.

so $tan\theta$ is,

$Tan\theta = \frac{a-1}{2\sqrt{a} }$

we get the answer of this question as  $Tan\theta = \frac{a-1}{2\sqrt{a} }$ .