Question

if 3 cotA= 4, check whether 1-tan^A/1+tan^A =cos^A- sin^A or not

Answer 1

Given 3 cotA = 4.

              cotA = 4/3. ----------------- (1).

We know that tan theta = 1/cot theta

                        tan A = 1/cotA

                                  = 1/4/3

                                  = 3/4.  --------------- (2)


We know that cosec^2 theta = 1 + cot^2 theta

                                                 = 1 + (4/3)^2

                                                 = 1 + 16/9

                                                 = 25/9.

                            cosec theta = 5/3.    ---------------- (3)



We know that cosec theta = 1/sin theta

                          (5/3) = 1/sin theta

                          5sin theta = 3 

                            sin theta = 3/5   ------------- (4)


We know that cos^2 theta = 1 - sin^2 theta

                                            = 1 - (3/5)^2

                                            = 1 - 9/25

                                            = 25 - 9/25

                                            = 16/25

                               cos A = 4/5.   ------------- (5)



Now, LHS:

$\frac{1 - tan^2A}{1 + tan^2A}$

$\frac{1 - (\frac{3}{4})^2 }{1 + ( \frac{3}{4})^2 }$

$\frac{1 - \frac{9}{16} }{1 + \frac{9}{16} }$

$\frac{16 - 9}{16 + 9}$

$\frac{7}{25}$.



Now, RHS: (4)&(5)

$cos^2A - sin^2A$

$( \frac{4}{5})^2 - ( \frac{3}{5} )^2$

$\frac{16}{25} - \frac{9}{25}$

$\frac{16-9}{25}$

$\frac{7}{25}$



LHS = RHS.


Hope this helps!

Answer 2

Answer:

Step-by-step explanation:

3cot A=4

TanA=1/4/3

=3/4...............1

Cosec^2=1+cot^2

=1+(4/3)^2

=1+16/9

=9+16/9

=25/9

Cosec^2 A=25/9

Degree of cosec goes that side and becomes root

CosecA=root of 25/9

=5/3

CosecA=3/9..............2

CosecA=1/sin theta

5/3=1/sin theta

5sin theta =3

Sin theta =3/5..............3

Cos^2 theta=1-sin^2

=q-(3/5)^2

=1-9/25

=25-9/25

=16/25

Cos^2=16/25

Square of cos becomes root on other side

Cos theta =root of 16/25

Cos theta =4/5.............. 4

=1-tan^2A/1+tan^A

1-(3/4)^2/1+(3/4)^2

=1-9/16/1+9/16

=16-9/16/16+9/16

=16-9/16+9

=7/25

Cos^2A-sin^2A

(4/5)^2-(3/5)^2

=16/25-9/25

16-9/25

=7/25

Therefore

LHS=RHS