Question

please solve it , don't give useless answers please .​

Answer 1

Step-by-step explanation:

i) Given,

a cos theta - b sin theta =c

squaring both sides,

(a cos theta - b sin theta)^2 = c^2

a^2cos^2 theta+ b^2 sin^2 theta - 2ab cos theta. sin theta = c^2

a^2 cos^2 theta + b^2 sin^2 theta - c^2

= 2ab cos theta.sin theta .................(1)

Now squaring LHS,

(a sin theta + b cos theta)^2

= a^2 sin^2 theta+b^2 cos^2 theta+2 ab cos theta.sin theta

(a sin theta+ b cos theta)^2

= a^2 sin^2 theta+b^2 cos^2 theta +( a^2 cos^2 theta+ b^2 sin^2 theta -c^2)

[from equation 1]

(a sin theta + b cos theta)^2

= a^2 (sin^2 theta+cos^2 theta)+ b^2(sin^2 theta+cos^2 theta) - c^2 [As sin^2 theta+cos^2 theta= 1]

(a sin theta + b cos theta)^2

= a^2+b^2-c^2

(a sin theta+ b cos theta) = ±√a^2 + b^2 - c^2

so, LHS= RHS

ii) Given,

tan^2 theta = 1-e^2 ........... (1)

[sec^2 theta= 1+tan^2 theta]

sec^2 theta= (1+1-e^2) [from 1]

sec^2 theta=(2-e^2)

sec theta=√(2-e^2) = (2-e^2)^1/2

Now taking LHS,

sec theta + tan^3 theta.cosec theta

= sec theta + tan^2 theta. tan theta.cosec theta

= sec theta + tan^2 theta.sec theta (As tan theta.cosec theta= sec theta)

= sec theta (1+tan^2 theta) ..............(2)

Putting the values in (2),

= (2-e^2)^1/2 . (2-e^2)^1

= (2-e^2)^3/2 [If bases are same in multiplication then powers are added]

so, LHS = RHS