Question

It sino + tano = m
tano - sino an
Then express the
values of m²-n² in terms
of M and N​

Answer 1

$\huge{\blue{\underline{\underline{\bf{Correct\;Question:-}}}}}$

If sin Ф + tan Ф = m and tan Ф - sin Ф = n. Then prove that m² - n² = 4√mn.

$\huge{\blue{\underline{\underline{\bf{Answer:-}}}}}$

$\sf{\implies \tan \theta + \sin \theta = m\;\;\;\;........(1)}$

$\sf{\implies \tan \theta - \sin \theta = n\;\;\;\;........(2)}$

Now, adding equation (1) & (2), we get

$\sf{\implies 2\tan \theta=m+n}$

$\sf{\implies \tan \theta = \dfrac{m+n}{2}}$

Now, substitute the value of tan Ф in equation (1), we get

$\sf{\implies \dfrac{m+n}{2}+\sin \theta = m}$

$\sf{\implies \sin \theta = m-\dfrac{m+n}{2}}$

$\sf{\implies \sin \theta = \dfrac{2m+-m-n}{2}=}$

$\sf{\implies \sin \theta = \dfrac{m-n}{2}}$

Now, we know that sec²Ф - tan²Ф = 1

$\sf{\implies \dfrac{1}{\cos^{2}\theta} - \tan^{2}\theta = 1}$

$\sf{\implies \dfrac{1}{1-\sin^{2}\theta}-\tan^{2}\theta = 1}$

Now, put the values of sin Ф & tan Ф from above.

$\sf{\implies \dfrac{1}{1-\Bigg(\dfrac{m-n}{2}\Bigg)^{2}}-\Bigg(\dfrac{m+n}{2}\Bigg)^{2}=1}$

$\sf{\implies \dfrac{1-\Bigg(\dfrac{m+n}{2}\Bigg)^{2}+\Bigg(\dfrac{m+n}{2}\Bigg)^{2}+\Bigg(\dfrac{m-n}{2}\Bigg)^{2}}{1-\Bigg(\dfrac{m-n}{2}\Bigg)^{2}} =1}$

$\sf{\implies 1-\Bigg(\dfrac{m+n}{2}\Bigg)^{2}+\Bigg[\dfrac{(m-n)(m+n)}{4}\Bigg]^{2}=1-\Bigg(\dfrac{m-n}{2}\Bigg)^{2}}$

$\sf{\implies \Bigg(\dfrac{m^{2}-n^{2}}{4}\Bigg)^{2}=\Bigg(\dfrac{m+n}{2}\Bigg)^{2} -\Bigg(\dfrac{m-n}{2}\Bigg)^{2}}$

$\sf{\implies \Bigg(\dfrac{m^{2}-n^{2}}{4}\Bigg)^{2}=\dfrac{m^{2}+n^{2}+2mn}{4}-\dfrac{m^{2}+n^{2}-2mn}{4}}$

$\sf{\implies \Bigg(\dfrac{m^{2}-n^{2}}{4}\Bigg)^{2}=\dfrac{m^{2}+n^{2}+2mn-m^{2}-n^{2}+2mn}{4}}$

$\sf{\implies \Bigg(\dfrac{m^{2}-n^{2}}{4}\Bigg)^{2} = \dfrac{4mn}{4}}$

$\sf{\implies \dfrac{m^{2}-n^{2}}{4}\Bigg=\sqrt{mn}}$

$\sf{\implies m^{2}-n^{2}=4\sqrt{mn}}$

Hence Proved!!!

Answer 2

Given :----

• sinA + tanA = m
• tanA - sinA = n

Question :---- we have to Prove m²-n² = 4√mn

Assume

sinA+tanA = m ------------------ Equation (1)

tanA - sinA = n ------------------ Equation (2)

Squaring and adding both Equations now or using ,

(a+b)² - (a-b)²= 4ab

we get,

m² - n² = 4tanAsinA -------------- [LHS]

Now, Lets solve RHS, by putting value of m & n ....

4√mn

→ 4√(tanA+sinA)(tanA-sinA)

→ 4√(tan²A - sin²A) [(a+b)(a-b) = (a²-b²)]

→ 4√[(sin²A/cos²A) - sin²A] [ tanA = sinA/cosA ]

$\implies4 \sqrt{ \frac{ \sin^{2}a - \sin^{2}a \times \cos^{2} a }{cos^{2} a} } \\ \\\implies 4 \sqrt{ \frac{sin^{2}a(1 - cos^{2} a)}{ \cos^{2} a } } \: \\ \\ \implies \: 4 \sqrt{ \frac{sin^{4} a}{cos ^{2} a} } \\ \\ \implies \: \frac{ 4sin^{2}a }{ {cos}a } \\ \\ \implies \: 4tana \times sina$

$\large\red{\boxed{\sf </strong><strong>LHS</strong><strong>=</strong><strong>RHS</strong><strong>}}$

Hence ,

$\large\</strong><strong>g</strong><strong>r</strong><strong>e</strong><strong>e</strong><strong>n</strong><strong>{\boxed{\sf </strong><strong>{m}^{2} - {n}^{2} = 4 \sqrt{mn}</strong><strong>}}$

[ Hope i Helped you in easiest way .]