Question

ii)
(1+cot theta+tan theta)(sin theta -cos theta)/
sec*3 0 -cosec*3theta
= sin*2 theta cos2theta​

Answer 1

Answer:

Explanation:1+cot theta +tan theta(sin theta -cos theta)/sec³theta -cosec³theta

converting all in the form of sin and cos

(1+cos theta /sin theta +sin theta /cos theta)(sin theta-cos theta)÷1/cos²theta -1/sin²theta

=[cos²theta+sin²theta+sin theta +cos theta/sin theta.cos theta]×(sin theta -cos theta) ÷sin³theta-cos³theta/sin³theta.cos3theta

=(1+sin theta.cos theta/sin theta cos theta)(sin theta.cos theta)÷sin theta-cos theta)(sin²theta +cos²theta+sin theta cos theta)/sin³theta.cos³theta

=1+sin theta cos theta/sin theta cos theta×sin theta-cos theta ×sin³theta.cos ³theta /sin theta-cos theta(1+sin theta.cos theta)

=sin²theta.cos²theta

Answer 2

Answer:

LHS =

(1+cot theta+ tan theta)(sin theta -cos theta)/

sec*3 0 -cosec*3theta

      ={ (1+cos theta/sin theta + sin theta/cos theta)(sin theta-cos theta) } / {1/cos3 theta - 1/sin3 theta}

$=(1+ (cos^{2} theta + sin^{2} theta/ cos theta*sin theta))(sin theta-cos theta)/ (sin^{3}theta- cos^{3}theta/sin^{3} theta.cos^{3} theta)$

$=(cos theta.sin theta +cos^{2}theta +sin^{2} sin theta)(sin theta-cos theta)(sin^{3} *cos^{3} )/( cos theta.sin theta ( sin theta-cos theta)(sin^{2}+cos^{2}+ sin theta.cos theta))$

$\frac{sin^{3} cos^{3}}{sin theta. cos theta} =\frac{sin theta .cos theta(sin^{2}theta. cos^{2}theta)}{sin theta .cos theta}$

$sin^{2} theta.cos^{2} theta$

= RHS

Hence proved!