Question

Prove both the above​

Answer 1

hii

here is u answer

first 2 photo is of 1 question and

another 2vphoto of 2 question

Answer 2

Note : x = θ

1.

LHS

$\to \sf \dfrac{1+cosx+sinx}{1+cosx-sinx}$

Divide num. and den. by cos x ,

$\to \sf \dfrac{\frac{1}{cosx}+\frac{cosx}{cosx}+\frac{sinx}{cosx}}{\frac{1}{cosx}+\frac{cosx}{cosx}-\frac{sinx}{cosx}}$

• sec x = 1 / cos x
• sin x / cos x = tan x

$\to \sf \dfrac{secx+1+tanx}{secx+1-tanx}$

• sec²x - tan²x = 1

$\to \sf \dfrac{secx+tanx+sec^2x-tan^2x}{secx+1-tanx}$

• a² - b² = ( a + b ) ( a - b )

$\to \sf \dfrac{secx+tanx+(secx+tanx)(secx-tanx)}{secx+1-tanx}\\\\\to \sf \dfrac{(secx+tanx)\cancel{(1+secx-tanx)}}{\cancel{1+secx-tanx}}$

$\to \sf secx+tanx\\\\\to \sf \dfrac{1}{cosx}+\dfrac{sinx}{cosx}\\\\\leadsto \sf \pink{\dfrac{1+sinx}{cosx}}\ \; \bigstar$

RHS

2.

LHS

$\to \sf \dfrac{cosx-sinx+1}{cosx+sinx-1}$

Divide num. and den. by sin x ,

$\to \sf \dfrac{\frac{cosx}{sinx}-\frac{sinx}{sinx}+\frac{1}{sinx}}{\frac{cosx}{sinx}+\frac{sinx}{sinx}-\frac{1}{sinx}}$

$\to \sf \dfrac{cotx-1+cscx}{cotx+1-cscx}$

• csc²x - cot²x = 1

$\to \sf \dfrac{cscx+cotx-(csc^2x-cot^2x)}{1-cscx+cotx}$

• a² - b² = ( a + b ) ( a - b )

$\\ \to \sf \dfrac{cscx+cotx-[(cscx+cotx)(cscx-cotx)]}{1-cscx+cotx}$

$\to \sf \dfrac{(cscx+cotx)(\cancel{1-cscx+cotx})}{\cancel{1-cscx+cotx}}$

$\sf \leadsto cscx+cotx\ \; \green{\bigstar}$

RHS