Question

(4) ( seco – coso)( coto + tano) = tan8 seco​

Answer 1

(secθ-cosθ)(cotθ+tanθ)

(1/cosθ-cosθ)(cosθ/sinθ+sinθ/cosθ)

[(1-cos²θ)/cosθ ] [(cos²θ+sin²θ)/sinθcosθ]

sin²θ/cosθ * 1/sinθcosθ

sinθ/cosθ*1/cosθ

tanθ*secθ

hence proved

Answer 2

Answer :-

$LHS = (sec\theta-cos\theta)(cot\theta+tan\theta)$

$=(sec\theta-cos\theta)(\frac{cos^{2}\theta + sin^{2}\theta}{sin\theta cos\theta} )$

$=(sec\theta-cos\theta)(\frac{cos\theta}{sin\theta} + \frac{sin\theta}{cos\theta} )$

$=(sec\theta-cos\theta)\frac{1}{sin\theta cos\theta}$

$=(\frac{1}{cos\theta} -\frac{cos\theta}{1} ) \frac{1}{sin\theta cos\theta}$

$=(\frac{1-cos^{2}\theta}{cos\theta}) \frac{1}{sin\theta cos\theta}$

$=\frac{sin^{2}\theta}{sin\theta cos^{2} \theta}$

$=\frac{sin\theta}{cos^{2}\theta}$

$=\frac{sin\theta}{cos\theta}$ × $\frac{1}{cos\theta}$

$=tan\theta sec\theta$

$RHS = tan\theta sec \theta$

$LHS=RHS$ $hence,$ $verified.$

☆ Hope it Helps ☆

@NightRead