Click here:point_up_2:to get an answer to your question :writing_hand:prove rightarrow sin 90othetacos 90othetadfrac tan theta1tan 2theta
Click here👆to get an answer to your question ✍️ Prove rightarrow sin -90-o-theta-cos -90-o-theta-dfrac -tan theta-1-tan -2-theta-
Prove sin-90-x2212-x3B8-cos-90-x2212-x3B8-tan-x3B8-1-tan2-x3B8-LHS-sin-90-x2212-x3B8-cos-90-x2212-x3B8-cos-x3B8-sin-x3B8-12-2sin-x3B8-cos-x3B8-12sin2-x3B8-12-2tan-x3B8-1-tan2-x3B8-tan-x3B8-1-tan2-x3B8-RHS
Prove rightarrow sin (90^{o}-theta)cos (90^{o}-theta)=dfrac {tan theta}{1+ tan ^{2}theta}
सिद्ध कीजिए (cos (90^(@)-A)sin (90^(@)-A))/(tan (90^(@)-A))=sin^(2)A.
सिद्ध कीजिए कि sin(90^(@)-theta)cos(90^(@)-theta)=(tan theta)/(1+cot^(2 )(90^(@)-theta)), 10
cos(90−θ)sec(90−θ)tanθ/cosec (90−θ)sin(90−θ)cot(90−θ)+tan(90−θ)/cotθ.
Prove that : (i) sinthetacos(90^(@)-theta)+sin(90^(@)-theta)costh
Evaluate without using trigonometric table: displaystyle frac{{cos e{c^2}left( {{{90}^ circ } - theta } right) - {{tan }^2}theta }}{{4left( {{{cos }^2}{{48}^ circ + }{{cos }^2}{{42}^ circ }} right)}} - frac{{2{{tan }^2}{{30}^ circ }{{
dfrac { cos({ 90 }^{ o }-theta )costheta }{ tantheta } +{ cos }^{ 2 }({ 90 }^{ o }-theta )=1
The Squeeze Theorem
How to solve 0 = - 3sin^2 theta - 2 + 4sin theta + cos^ 2 theta for 0 <= theta < 2pi - Quora