Prove the following-cos -x cos -x-sin -x cos-2-2- 2 x

We have L.H.S. = cos(π+x)cos( x)/sin(π x) cos ( π/2+2) = cot^2 x = cos x.cos x/sin x( sin x) [ ∵cos(π+0)= cos θ , cos ( θ) = cos θ; sin(π θ)= sin θ, c ...

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Standard XI

Mathematics

Trigonometric Ratios for Sum of Two Angles

Prove the fol...

Question

Prove the following:

= $\frac{cos (π + x) cos (- x)}{sin (π - x) c o s (\frac{π}{2} + 2)} = c o t^{2} x$

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Solution

We have

L.H.S. = $\frac{cos (π + x) cos (- x)}{sin (π - x) c o s (\frac{π}{2} + 2)} = c o t^{2} x$

= $\frac{-cos x . cos x}{sin x (- sin x)}$

$⎡ ⎣ \begin{matrix} ∵ cos (π + 0) = - cos θ, cos (- θ) = cos θ sin (π - θ) = sin θ, cos (\frac{π}{2} + θ) = - sin θ \end{matrix} ⎤ ⎦$

= $\frac{- {cos}^{2} x}{- {sin}^{2} x} = {cot}^{2} x$ = R.H.S.

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Similar questions

Q. Prove the following:

\frac{cos (π + x) cos (- x)}{sin (π - x) cos (\frac{π}{2} + x)} = {cot}^{2} x

Q. Prove that

\frac{cos (π + x) cos (- x)}{sin (π - x) cos (\frac{π}{2} + x)} = {cot}^{2} x

Q. Prove the following:

\frac{c o s (π + x) c o s (- x)}{s i n (π - x) c o s (\frac{π}{2} + x)} = c o t^{2} x

Q. Prove

\frac{cos (π + x) cos (- x)}{sin (π - x) cos (\frac{π}{2} + x)} = {cot}^{2} x

[cos(π+x) cos(-x)] divided by / sin(π-x) cos(π/2 + x) = cot^2x

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Prove the following: = cos(π+x)cos(−x)sin(π−x)cos(π2+2)=cot2 x

We have L.H.S. = cos(π+x)cos(−x)sin(π−x)cos(π2+2)=cot2 x =-cosx.cosxsinx(−sin x) ⎡⎣∵cos(π+0)=−cos θ,cos(−θ)=cos θsin(π−θ)=sin θ,cos(π2+θ)=−sin θ⎤⎦ = −cos2 x−sin2 x=cot2 x= R.H.S.

Prove the following:

= $\frac{cos (π + x) cos (- x)}{sin (π - x) c o s (\frac{π}{2} + 2)} = c o t^{2} x$