Question

if sec tetha+tan tetha=p, prove that sin tetha=p²-1/p²+1​

Answer 1

Answer:

Step-by-step explanation:

$\Large{\underline{\rm{Given:}}}$

• sec θ + tan θ = p

$\Large{\underline{\rm{To\:Prove:}}}$

$\sf{sin\:\theta=\dfrac{p^{2} -1}{p^{2}+1 }}$

$\Large{\underline{\rm{Identities\:used:}}}$

(a + b)² = a² + b² + 2ab

sec²θ - 1 = tan²θ

tan²θ + 1 = sec²θ

tan θ = sinθ/cos θ

1/ sec θ = cos θ

$\Large{\underline{\rm{Proof:}}}$

We know that sec θ + tan θ = p.

We have to prove that,

$\sf{sin\: \theta=\dfrac{p^{2} -1}{p^{2}+1 }}$

Taking the RHS of the above equation,

$\sf{\dfrac{p^{2} -1}{p^{2}+1 }}$

Substitute the value of p from given,

$\sf{\dfrac{(sec\theta+tan\theta)^{2} -1}{(sec\theta+tan\theta)^{2}+1 }}$

Expanding by using identities,

$\sf{\dfrac{(sec^{2} \theta+tan^{2} \theta+2\:sec\theta\:tan\theta) -1}{(sec^{2} \theta+tan^{2} \theta+2\:sec\theta\:tan\theta)+1 }}$

Again applying identities,

$\sf{\dfrac{tan^{2} \theta+tan^{2} \theta+2\:sec\theta\:tan\theta}{sec^{2} \theta+sec^{2} \theta+2\:sec\theta\:tan\theta }}$

$\sf{\dfrac{2tan^{2} \theta+2\:sec\theta\:tan\theta}{2sec^{2} \theta+2\:sec\theta\:tan\theta }}$

Taking 2 tanθ common from numerator and 2 secθ common from denominator,

$\sf{\dfrac{2tan\theta(tan\theta+sec\theta)}{2sec \theta(sec\theta+tan\theta) }}$

Cancelling tanθ + sec theta and 2 on both numerator and denominator,

$\sf{\dfrac{tan\theta}{sec\theta} =\dfrac{sin\theta}{cos\theta}\times \dfrac{1}{sec\theta} }$

= sin θ

Hence LHS = RHS

Hence proved.

Answer 2

$\Large{\underline{\rm{Given:}}}$

sec θ + tan θ = p

$\Large{\underline{\rm{To\:Prove:}}}$

$\sf{sin\:\theta=\dfrac{p^{2} -1}{p^{2}+1 }}$

$\Large{\underline{\rm{Identities\:used:}}}$

(a + b)² = a² + b² + 2ab

sec²θ - 1 = tan²θ

tan²θ + 1 = sec²θ

tan θ = sinθ/cos θ

1/ sec θ = cos θ

$\Large{\underline{\rm{Proof:}}}$

We know that sec θ + tan θ = p.

We have to prove that,

$\sf{sin\: \theta=\dfrac{p^{2} -1}{p^{2}+1 }}$

Taking the RHS of the above equation,

$\sf{\dfrac{p^{2} -1}{p^{2}+1 }}$

Substitute the value of p from given,

$\sf{\dfrac{(sec\theta+tan\theta)^{2} -1}{(sec\theta+tan\theta)^{2}+1 }} </p><p>$

Expanding by using identities,

$\sf{\dfrac{(sec^{2} \theta+tan^{2} \theta+2\:sec\theta\:tan\theta) -1}{(sec^{2} \theta+tan^{2} \theta+2\:sec\theta\:tan\theta)+1 }}$

Again applying identities,

$\sf{\dfrac{tan^{2} \theta+tan^{2} \theta+2\:sec\theta\:tan\theta}{sec^{2} \theta+sec^{2} \theta+2\:sec\theta\:tan\theta }}$

$\sf{\dfrac{2tan^{2} \theta+2\:sec\theta\:tan\theta}{2sec^{2} \theta+2\:sec\theta\:tan\theta }}$

Taking 2 tanθ common from numerator and 2 secθ common from denominator,

$\sf{\dfrac{2tan\theta(tan\theta+sec\theta)}{2sec \theta(sec\theta+tan\theta) }} </p><p>$

Cancelling tanθ + sec theta and 2 on both numerator and denominator,

$\sf{\dfrac{tan\theta}{sec\theta} =\dfrac{sin\theta}{cos\theta}\times \dfrac{1}{sec\theta} }$

Hence LHS = RHS

Hence proved.