Question

Here Is A Question For You Guys ♥️♥️

Please Help Me In This Question...

Simplest Answer Will Be Brainlest ✌️✌️☺️

@Shona...❣️❣️​

Answer 1

( 1 + cot Ф + tan Ф )( sin Ф - cos Ф )

⇒ ( 1 + cos Ф / sin Ф + sin Ф / cos Ф )( sin Ф - cos Ф )

⇒ ( sin Ф cos Ф + sin²Ф + cos²Ф )( sin Ф - cos Ф ) / sin Ф cos Ф

Use the expansion : ( a - b )( a² + ab + b² ) = a³ - b³

⇒ ( sin³Ф - cos³Ф )/ sin Ф cos Ф

So ( sin³Ф - cos³Ф ) / ( sec³ - cos³Ф ) ( sin Ф cos Ф )

⇒ ( 1/cosec³Ф - 1/sec³Ф ) / ( sec³Ф - cos³Ф ) ( sinФ cosФ )

⇒ ( sec³Ф - cos³Ф )/sec³Фcosec³Ф ) / ( sec³Ф - cos³Ф )( 1/cosecФ.1/secФ )

⇒ secФcosecФ / sec³Фcosec³Ф

⇒ 1 / sec²Фcosec²Ф

⇒ sin²Фcos²Ф

Answer 2

Step-by-step explanation:

Note: For the better understanding I am replacing θ with A.

$Given:\frac{(1+cotA+tanA)(sinA-cosA)}{sec^3A-cosec^3A}$

$=\frac{1+(\frac{cosA}{sinA}+\frac{sinA}{cosA})(sinA-cosA)}{(\frac{1}{cos^3A}-\frac{1}{sin^3A})}$

$=\frac{(\frac{sinAcosA+cos^2A+sin^2A}{sinAcosA})*(sinA-cosA)}{\frac{sin^3A-cos^3A}{cos^3Asin^3A}}$

$=\frac{(\frac{sinAcosA+1}{sinAcosA})(sinA-cosA)}{\frac{(sinA-cosA)(sin^2A+cos^2A+sinAcosA}{cos^3Asin^3A})}$

$=\frac{(sinAcosA+1)(sinA-cosA)}{sinAcosA} * \frac{cos^3Asin^3A}{(sinA-cosA)(1+sinAcosA)}$

$=\frac{cos^3Asin^3A}{sinAcosA}$

$=\boxed{sin^2Acos^2A}$

Hope it helps!