Question

Hi geniuses. For u.

Answer 1

(1) sin A + sin B = 2

0 <  A < 90     ,      0 < B < 90

sin A < 1

sin B <  1

Their maximum values can be 1

So sin A and sin B must be 1 . Only then :

sin A + sin B = 2

So : sin A = sin B = 1

A = B = 90°

cos A + cos B = cos 90 + cos 90

                       = 0 + 0

                       = 0

( Answer ) 0

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(3) $sin^6\theta + cos^6\theta + 3 sin^2\theta cos^2\theta$

$\implies (sin^2\theta)^3+(cos^2\theta)^3 + 3 sin^2\theta cos^2\theta$

[ Using : $a^3+b^3=(a+b)(a^2-ab+b^2)$

$\implies (sin^2\theta+cos^2\theta)(sin^4\theta - sin^2\theta cos^2\theta+cos^4\theta)+3sin^2\theta cos^2\theta$ ]

We know that : $sin^2\theta +cos ^2\theta = 1$

$\implies sin^4\theta-sin^2\theta cos^2\theta+ cos^4\theta+3sin^2\theta cos^2\theta$

$\implies sin^4\theta + 2sin^2\theta cos^2\theta + cos^4\theta$

We know that : $(a+b)^2=a^2+b^2+2ab$

$\implies (sin^2\theta+cos^2\theta)^2$

$\implies 1^2$

$\implies 1$

The value will be 1.

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(4) cosec²θ = 2 cot θ

==> 1/sin²θ = 2 cot θ [ cosec²θ = 1/sin²θ ]

==> 1/sin²θ = 2 cosθ / sin θ [ cot θ = cos θ/sin θ ]

==> 1/sin θ = 2 cos θ

==> cos θ sin θ = 1/2

By trial and error method

==> cos θ cos ( 90 - θ ) = 1/2

==> cos θ cos ( 90  - θ ) = 1/(√2) × 1/(√2)

==> cos θ  cos ( 90 - θ ) = cos 45 × cos 45

θ = 45 .

or , 90 - θ = 45

==> θ = 90 - 45 = 45

So θ must be 45

$\boxed{\bf{ANSWERS}}$

(1) 0

(3) 1

(4) 45°

You said that you do not require 2 but if you ask I may do it :)

Hope it helps u :-)

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