Question

(sinq+cosq) (tanq+cotq) = secq+cosecq

Answer 1

(sinΘ+cosΘ)(tanΘ+cotΘ) 
as we know tan Θ = sinΘ/cosΘ  and cot Θ = cos Θ/sinΘ lets apply that

(sinΘ+cosΘ)(sinΘ/cosΘ + cosΘ/sinΘ)
(sinΘ+cosΘ)(sin^2Θ + cos ^2Θ / sin Θ x cos Θ) 
(sinΘ+cosΘ)( 1 / sin Θ x cos Θ) ( because sin^2Θ + cos^2Θ = 1) 
(sinΘ+cosΘ)/( sin Θ x cos Θ) 
sinΘ/ sin Θ x cos Θ + cosΘ/ sin Θ x cos Θ ( distribute property)
1/cosΘ + 1/sinΘ
secΘ+cscΘ = RHS

Answer 2

COT tita+TAN tita=COSEC tita.SEC tita

Step-by-step explanation:

SOLUTION:

L.H.S.=COT tita +TAN tita

=COS tita\SIN tita+SIN tita\COS tita

.....[COT tita=COS tita\SIN tita

TAN tita =SIN tita\COS tita]

=COS²tita+SIN²tita\SIN tita. COS tita

= 1\SIN tita. COS tita

....(therefore SIN²tita+cos²tita=1)

COSEC tita. SEC tita

L.H.S.=R.H.S.

hence COT tita+TAN tita=COSEC tita. SEC tita