Question

If m = tanA + sinA and n = tanA - sinA, then express (m² - n²)² in terms of m and n.

Options are given below.

(A) 16mn

(B) 4mn

(C) 32mn

(D) 8mn

Answer 1

Answer:

$\bold\red{(A)16mn}$

Step-by-step explanation:

Given,

m = tanA + sinA

squaring both sides, we get

$=>{m}^{2}={(tanA+sinA)}^{2}\\\\=>{m}^{2}={tan}^{2}A+{sin}^{2}A+2tanAsinA\:......(i)$

Also,

n = tanA - sinA

squaring both sides, we get

$=>{n}^{2}={(tanA-sinA)}^{2}\\\\=>{n}^{2}={tan}^{2}A+{sin}^{2}A-2tanAsinA\:.....(ii)$

Subtracting eqn (ii) from (ii),

we get,

$=>{m}^{2}-{n}^{2}=4tanAsinA$

Again, squaring both sides, we get

$=>{({m}^{2}-{n}^{2})}^{2}={4tanAsinA}^{2}=16{tan}^{2}A{sin}^{2}A$

But, ${sin}^{2}A=1-{cos}^{2}A$

so,

putting the value, we get

$=>{({m}^{2}-{n}^{2})}^{2}=16{tan}^{2}A(1-{cos}^{2}A)$

$=>{({m}^{2}-{n}^{2})}^{2}=16({tan}^{2}A-{sin}^{2}A)$

But, ${tan}^{2}A-{sin}^{2}A=mn$

Therefore,

$=>{({m}^{2}-{n}^{2})}^{2}=16mn$

Hence,

Option (A)16mn is correct answer.

Answer 2

Question :

If m = tan A + sin A and n = tan A - sin A , then express (m² - n²)² in terms of m and n ,

1. 16 mn
2. 4 mn
3. 32 mn
4. 8 mn

Solution :

Given ,

m = tan A + sin A

➠ m² = (tan A + sin A)²

➠ m² = tan²A + sin²A + 2 tan A.sin A

n = tan A - sin A

➠ n² = (tan A - sin A)²

➠ n² = tan²A + sin²A - 2 tan A.sin A

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m² - n²

➠ tan²A + sin²A + 2 tan A.sin A - ( tan²A + sin²A - 2 tan A.sin A )

➠ tan²A + sin²A + 2 tan A.sin A - tan²A - sin²A + 2 tan A.sin A

➠ 4 tan A . sin A

So , m² - n² = 4 tan A . sin A

Let's calculate (m² - n²)²

$:\implies \sf (4\ tan\ A\ .\ sin\ A)^2$

$:\implies \sf 16\ tan^2A\ .\ sin^2A$

• sin²A = 1 - cos²A

$:\implies \sf 16\ tan^2A(1-cos^2A)$

$:\implies \sf 16\ (tan^2A\ - tan^2A.cos^2A)$

$:\implies \sf 16\ (\ tan^2A-sin^2A\ )$

• a² - b² = (a + b)(a - b)

$:\implies \sf 16\ (tan\ A+sin\ A)(tan\ A-sin\ A)$

$:\implies \sf 16\ (m)(n)$

$:\implies \sf \pink{16mn}\ \; \bigstar$

Option (1)

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