Question

hey mates after a long time plzz solve my question thanx I know the writing is not ohk but plzz

Answer 1

Heya mate, Here is ur answer

Refer to the attachment ☺

Identities used -

(a+b)^2 = a^2+b^2+2ab

Sin^2 A + cos ^2 A =1

Sin^2 A = 1- cos ^2 A


Steps-

1 ) Simplify P .

2) Solve p^2 .

3) Put the values in the question.

4) Take the LCM.

5) Use the identities that I wrote above.

6) Solve further.

7) You will get ur answer.☺


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Warm regards

@Laughterqueen

Be Brainly ✌✌✌

Answer 2

Answer:

Sec A + Tan A = p

To Prove: Sin A = ( p² - 1 / p² + 1 )

Proof:

Calculating p² - 1 we get,

⇒ ( Sec A + Tan A )² - 1

⇒ ( Sec²A + Tan²A + 2 SecA.TanA - 1 )

⇒ ( Sec²A - 1 + Tan²A + 2 SecA.TanA )

We know the identity that, Sec²A - 1 = Tan²A. Substituting that we get,

⇒ ( Tan²A + Tan²A + 2SecA.TanA )

⇒ ( 2 Tan²A + 2 SecA.TanA )

Taking 2 TanA common we get,

⇒ 2 TanA ( Tan A + Sec A )   ... ( Equation 1 )

Now solving the denominator ( p² + 1 ) we get,

⇒ ( Sec A + Tan A )² + 1

⇒ ( Sec²A + Tan²A + 2SecA.TanA + 1 )

We know that, 1 + Tan²A = Sec²A. Hence substituting that we get,

⇒ ( 1 + Tan²A + Sec²A + 2 SecA.TanA )

⇒ ( Sec²A + Sec²A + 2 SecA.TanA )

⇒ ( 2 Sec²A + 2 SecA.TanA )

Taking 2 SecA common we get,

⇒ 2 SecA ( Sec A + Tan A ) .... ( Equation 2 )

Now dividing Equation 1 by Equation 2 we get,

$\implies \dfrac{ 2\: TanA\: ( Tan A + Sec A ) }{ 2\: SecA \:( Sec A + Tan A ) }$

( Tan A + Sec A ) will get cancelled and we get,

$\implies \dfrac{ 2\: Tan A }{2 \: Sec A} \implies \dfrac{Tan A}{SecA}$

Writing both of them in terms of SinA and CosA we get,

$\implies \dfrac{\dfrac{SinA}{CosA}}{ \dfrac{1}{CosA} }\\\\\\\\\implies \dfrac{SinA}{CosA} \times \dfrac{CosA}{1} \\\\\\\text{CosA gets cancelled and we get,}\\\\\implies \dfrac{SinA}{1} = SinA$

Hence Proved !!