Question

Prove the following :
(i)sin θ sin (90° − θ) − cos θ cos (90° − θ) = 0
(ii)$\frac{cos(90^{0}-\Theta)sec(90^{0}-\Theta)tan\Theta}{cosec(90^{0}-\Theta)sin(90^{0}-\Theta)cot(90^{0}-\Theta)}+\frac{tan(90^{0}-\Theta)}{cot\Theta} =2$
(iii)$\frac{tan(90^{0}-A)cotA}{cosec^{2}A}-cos^{2}A=0$
(iv)tex]\frac{cos(90^{0}-A)sin(90^{0}-A)}{tan(90^{0}-A)}=sin^{2} A[/tex]
(v)sin (50° − θ) − cos (40° − θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1

Answer 1

SOLUTION :  

Given :  

(i) sin θ sin (90° − θ) − cos θ cos (90° − θ) = 0

L.H.S  

sin θ sin (90° − θ) − cos θ cos (90° − θ)

= sin θ. cos θ − cos θ . sinθ

[sin (90° − θ) = cos θ & cos (90° − θ) = sin θ]

= sin θ [cos θ -  cos θ]

= sin θ × 0

sin θ sin (90° − θ) − cos θ cos (90° − θ) = 0  

L.H.S = R.H.S

Hence, proved

(iv)  cos (90° − A) . sin (90° − A) / tan (90° − A)  = sin²A

L.H.S

cos (90° − A) . sin (90° − A) / tan (90° − A)

= sinA . cos A/ cotA

[sin (90° − θ) = cos θ & cos (90° − θ) = sin θ,tan (90° −  θ ) = cot θ]

= sinA . cos A / (cos A/ sin A)

= sinA . cos A × sin A /  cos A

cos (90° − A) . sin (90° − A) / tan (90° − A) = sin²A

L.H.S = R.H.S

Hence, proved

SOLUTION (ii), (iii) & (v) are in the attachment.

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Answer 2

<b></b>HeYaA mAtE

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