Question

26 both please help me geniuses??

Answer 1

please mark it brainliest

Answer 2

Answer:

secθ + cosecθ

Step-by-step explanation:

(a)

Given: (sinθ + cosθ)(tanθ + cotθ)

= (sinθ+cosθ)[sinθ/cosθ + cosθ/sinθ]

= (sinθ + cosθ)[sin²θ + cos²θ/sinθ cosθ]

= (sinθ + cosθ)[1/sinθ cosθ]

= (sinθ + cosθ)/(sinθ cosθ)

= (sinθ/sinθ cosθ) + (cosθ/sinθ cosθ)

= (1/cosθ) + (1/sinθ)

= secθ + cosecθ

(b)

Here, I am writing θ as A.

(i)

$Given:\frac{sinA-cosA+1}{sinA+cosA-1} * \frac{sinA+cosA+1}{sinA+cosA+1}$

$=\frac{sin^2A+sinAcosA+sinA-sinAcosA-cos^2A-cosA+sinA+cosA+1}{(sinA+cosA)^2-1}$

$=\frac{2sin^2A+2sinA}{sin^2A+cos^2A+2sinAcosA-1}$

$=\frac{2sinA(1+sinA)}{2sinAcosA}$

$=\frac{1+sinA}{cosA}$

$=\frac{1+sinA}{cosA}*\frac{1-sinA}{1-sinA}$

$=\frac{1-sin^2A}{cosA(1-sinA)}$

$=\frac{cos^2A}{cosA(1-sinA)}$

$=\frac{cosA}{1-sinA}$

(ii)

$Given:\frac{1}{secA-tanA}$

$=\frac{1}{\frac{1}{cosA}-\frac{sinA}{cosA}}$

$=\frac{cosA}{1-sinA}$

∴ (i) = (ii) = 1/secA - tanA

Hope it helps!