Question

if 3 cot A=4 ,check whether 1-tan square A /1+ tan square A = cos square A - sin square A or not...​

Answer 1

Answer:

${\blue{\underline{\underline{\bold{\huge{Given:-}}}}}}$

3 cot A = 4

=> cot A = 4/3

${\purple{\underline{\underline{\bold{\huge{To Prove:-}}}}}}$

1-tan square A /1+ tan square A = cos square A - sin square A

${\pink{\underline{\underline{\bold{\huge{Solution:-}}}}}}$

Since, cot A = 4/3 {cot Ѳ = base/perpendicular}

Let the base be 4x and perpendicular be 3x

According to Pythagoras Theorem:

(hypotenuse)² = (perpendicular)² + (base)²

=> (hypotenuse)² = (4x)² + (3x)²

=> (hypotenuse)² = 16x² + 9x²

=> (hypotenuse)² = 25x²

=> hypotenuse = √25x²

=> hypotenuse = 5x

So,

tan A = perpendicular/base = 3x/4x = ¾

cos A = base/hypotenuse = 4x/5x = ⅘

sin A = perpendicular/hypotenuse = 3x/5x = ⅗

L.H.S = 1 - tan² A / 1 + tan² A

$= \frac{1 - { (\frac{3}{4}) }^{2} }{1 + (\frac{3}{4}) ^{2} } \\ = \frac{1 - \frac{9}{16} }{1 + \frac{9}{16} } \\ = \frac{ \frac{16 - 9}{16} }{ \frac{16 + 9}{16} } \\ = \frac{ \frac{7}{16} }{ \frac{25}{16} } = \frac{7}{16} \times \frac{16}{25} = \frac{7}{25}$

R.H.S = cos² A - sin² A

${ (\frac{4}{5}) }^{2} - {( \frac{3}{5} )}^{2} \\ = \frac{16}{25} - \frac{9}{25} \\ = \frac{16 - 9}{25} \\ = \frac{7}{25}$

L.H.S = 7/25 , R.H.S = 7/25

=> L.H.S = R.H.S

=> 1 - tan² A / 1 + tan² A = cos² A - sin² A

$\huge\mathfrak{\pink{\fbox{\purple{\bigstar{\mathfrak{\green{Hence\; Proved}}}}}}}$

Answer 2

Answer:

cosA−sinA+1cosA+sinA−1

=(cosA−sinA+1)(cosA+sinA+1)(cosA+sinA−1)(cosA+sinA+1)

=(cosA+1¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯−sinA)(cosA+1¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯+sinA)(cosA+sinA¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯−1)(cosA+sinA¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯+1)

=(cosA+1)2−(sinA)2(cosA+sinA)2−12

=cos2A+2cosA+1−sin2Acos2A+sin2A+2sinAcosA−1

=cos2A+2cosA+cos2A1+2sinAcosA−1(∵ cos2A+sin2A=1)

=2cos2A+2cosA2sinAcosA

=2cos2A2sinAcosA+2cosA2sinAcosA

=cosAsinA+1sinA

=cotA+cosecA

=cosecA+cotA

Proved.