Question

tan square A minus 10 square b is equal to cos square B minus cos square A by cos square A into cos square B​

Answer 1

Correct Order of Question: tan²A - tan²B = (cos²B - cos²A)/cos²A cos²B

(Please write correct order of question to increase chances of getting valid answers)

L.H.S → tan²A - tan²B

→ sin²A/cos²A - sin²B/cos²B

→ (sin²A cos²B - sin²B cos²A)/cos²A cos²B [L.C.M]

By replacing sin²A by 1 - cos²A and sin²B by 1 - cos²B we get

→ [(1 - cos²A)cos²B - (1 - cos²B)cos²A]/cos²A cos²B

→ [cos²B - cos²A cos²B - cos²A + cos²A cos²B]/cos²A cos²B

→ (cos²B - cos²A)/cos²A cos²B = R.H.S

Hence Proved

Answer 2

Answer:

$\bold\red{To\:prove\colon}$

$\rm {tan}^{2}A - {tan}^{2}B =\dfrac{{Cos}^{2} B - {Cos}^{2} A }{{Cos}^{2} A {Cos}^{2}B }$

$\rule{300}{2}$

$\bold\red{Identity\:used:-}$

$\bullet$ $\bold\green{{tan}^{2}\theta=\dfrac{{Sin}^{2}\theta}{{Cos}^{2}\theta}}$

$\bullet$ $\bold\green{{Sin}^{2}\theta = 1-{Cos}^{2}\theta }$

$\rule{300}{1}$

$\bold\red{Proof↓}$

$\rm L.H.S. = {tan}^{2} A - {tan}^{2} B \\\\\rm \: \: \: \: = \dfrac{{Sin}^{2} A }{{Cos}^{2} A} - \dfrac{{Sin}^{2} B }{{Cos}^{2} B } \\\\\rm \: \: \: \: =\dfrac{{Sin}^{2} A {Cos}^{2} B - {Sin}^{2} B {Cos}^{2} A}{{Cos}^{2} A {Cos}^{2} B}\\\\\rm \: \: \: \: =\dfrac{(1-{Cos}^{2} A) {Cos}^{2} B - (1- {Cos}^{2} B) {Cos}^{2} A}{ {Cos}^{2} A {Cos}^{2} B} \\\\\rm \: \: \: \: =\dfrac{{Cos}^{2} B - \cancel{ {Cos}^{2} A {Cos}^{2} B}- {Cos}^{2} A + \cancel{{Cos}^{2} A {Cos}^{2} B} }{ {Cos}^{2} A {Cos}^{2} B}\\\\\rm\green \: \: \: \: = \dfrac{ {Cos}^{2} B - {Cos}^{2} A }{ {Cos}^{2} A {Cos}^{2} B}\\\\\rm\: \: \: \: =R.H.S.\: \:[Proved]$

$\therefore\rm L.H.S. = R.H.S. [Proved]$