Question

sin Q +secQ - cosecQ =tanQ +cotQ ​

Answer 1

Step-by-step explanation:

(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

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Answer 2

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