Question

given 4 Sin A =3 Cos A find the value of(a) SinA (b)CosA (c)Cot square A-cosec square A​

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

I hope everyone is familiar with the equation , cosec^2(A)-cot^2(A)=1.

i am slightly modifying it . i see this has a^2-b^2=(a+b)*(a-b).

so, we can write (cosecA+cotA)*(cosecA-cotA)=1.

now we know that cosecA+cotA=3 ( given)

so we can write, 3*(cosecA-cotA)=1. -----(i)

we know that cosecA=1/sinA and cotA=cosA/sinA.

so (i) becomes... (1/sinA)-(cosA/sinA)=1/3.

(1-cosA)/sinA=1/3.

(1-cosA)=sinA/3. ----------(ii)

we know that sin ^2(A)+cos^2(A)=1.

from which we get , sinA=(1-cos^2(A))^(1/2).

so (ii) becomes ... 3*(1-cosA)=(1-cos^2(A))^(1/2).

squaring both sides, we get

9*(1-cosA)^2=(1-cos^2(A)).

9*(1+cos^2(A)-2*cosA)=1-cos^2(A).

9+9*cos^2(A)-18*cosA=1-cos^2(A).

on rearranging we get ,

10cos^2(A)-18*cosA+8=0.--------(iii)

here , we get a quadratic equation of cosA .on solvin of which we get value of cosA.

since it is a quadratic equation we get two values of cosA .

on solving (iii) we get cos A= 1 or 4/5.

i hope you understood.