Class 11th

$\mathrm{28.}$$$$$ L.H.S = $L.H.S =cos\left(\frac{3\pi }{4}+x\right)-cos\left(\frac{3\pi }{4}-x\right)$

Using cos (A + B) = cos A cos B – sin A sin B

and cos (A – B) = cos A cos B + sin A sin B

$L.H.S =\left[cos\frac{3\pi }{4}cosx-sin\frac{3\pi }{4}sinx\right]-\left[cos\frac{3\pi }{4}cosx+sin\frac{3\pi }{4}sinx\right]$

$=cos\frac{3\pi }{4}cosx-sin\frac{3\pi }{4}sinx-cos\frac{3\pi }{4}cosx-sin\frac{3\pi }{4}sinx.$

$=-2sin\frac{3\pi }{4}sinx.$

$=-2sin\left(\frac{4\pi -\pi }{4}\right)sinx.$

27. L.H.S$L.HS=sin(x+1)xsin(x+2)x+cos(x+1)xcos(x+2)x.$

$=cos(x+1)xcos(x+2)x+sin(x+1)xsin(x+2)x.$

Let A = (n+1)x and B = (n+2)x

So, L.H.S = cosAcosB + sin Asin B

$=cos(A-B)$

Putting values of A and B we get,

L.H.S = $L.H.S = cos[(n+1)x-(n+2)x]$

$=cos[nx+x-nx-2x]$

$=cos(-x)$

$=cosx= R.H.S$ R.H.S

26. $$L.H.S $L.H.S.=cos\left(\frac{3\pi }{2}+x\right)cos(2\pi +x)\left[cot\left(\frac{3\pi }{2}-x\right)+cot(2\pi +x)\right]$

Here,

$cos\left(\frac{3\pi }{2}+x\right)=cos\left(\frac{4\pi -\pi }{2}+x\right)=cos\left(2\pi -\frac{\pi }{2}+x\right)$

$=cos\left[2\pi -\left(\frac{\pi }{2}-x\right)\right];VI quadrant.$

$cos\left(\frac{\pi }{2}-x\right);i I^{st} quadrant.$

$=sinx$

$cos(2x+x)=cosx,as xlies in I^{st}quadrant.$

$cot(2\pi +x)=cotx;asxliesin{I}^{st}quadrant.$

$cot\left(\frac{3\pi }{2}-x\right)=cot\left[\frac{4\pi -\pi }{2}-x\right]$

$=cot\left[2\pi -\frac{\pi }{2}-x\right]$

$=cot\left[2x-\left(\frac{\pi }{2}+x\right)\right];V{I}^{th} quadrant.$

$=-cot\left(\frac{\pi }{2}+x\right);I{I}^{nd}quadrat.$

$=-(-tanx)$

$=tanx.$

So.L.H.S $L.H.S=sinxcosx[tanx+cotx]$

$=sinxcosx\left[\frac{sinx}{cosx}+\frac{cosx}{sinx}\right]$

$=sinxcosx\frac{\left(si{n}^{2}x+co{s}^{2}x\right)}{cosxsinx}$

$=si{n}^{2}x+co{s}^{2}x$

= 1

= R.H.S.

23. (i) sin 75°= sin (45°+30°)

Using sin (x + y)= sin x cos y + cos x sin y we can write

22.  (i) sin 75°= sin (45°+30°)

Using sin (x + y)= sin x cos y + cos x sin y we can write

sin 75°

= sin 45°cos 30°+ 45° sin 30°