Question

If cot theta =3/4 prove that under root(sec theta - cosec theta/sec theta +cos theta) = 1/root7

plz ans fast

Answer 1

HELLO DEAR,

it seems questions have some error,

the correct question is if cotФ = 3/4 then prove that, $\bold{\sqrt{\frac{secФ - cosecФ}{secФ + cosecФ}}}$ = $\bold{1/\sqrt{7}}$

GIVEN:- cotФ = 3/4

[ figure is in the attachment]

IN ∆ABC, <B = 90°

so, cotФ = 3/4 = base/height = BC/AB

hence, BC = 3 , AB = 4

now, [by Pythagoras theorem],

AC² = AB² + BC²

AC² = 4² + 3²

AC² = 16 + 9

AC = √25

AC = 5

∵ AC = 5 , BC = 3 , AB = 4.

now, IN triangle, ABC

sinФ = AB/AC, cosФ = BC/AC

SinФ = 4/5 , cosФ = 3/5

if sinФ = 4/5 ⇒cosecФ = 5/4

if cosФ = 3/5 ⇒secФ = 5/3.

now, L.H.S = $\bold{\sqrt{\frac{secФ - cosecФ}{secФ + cosecФ}}}$

$\bold{=\sqrt{\frac{5/3 - 5/4}{5/3 + 5/4}}}$

$\bold{=\sqrt{\frac{5/12}{35/12}}}$

$\bold{=\sqrt{\frac{5}{35}}}$

$\bold{=\sqrt{1/7}}$

$\bold{=1/\sqrt{7}}$

Hence, $\boxed{\bold{\sqrt{\frac{secФ - cosecФ}{secФ + cosecФ}}}= \bold{1/\sqrt{7}}}$

thus, L.H.S = R.H.S.

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

cotθ=

4

3

⇒tanθ=

3

4

⇒sinθ=

5

4

,cosθ=

5

3

cscθ=

4

5

,secθ=

3

5

from Pythagorean triplets

Given,

secθ+cscθ

secθ−cscθ

=

3

5

+

4

5

3

5

−

4

5

=

35

12

12

5

=

7

1