Question

( cosescA - sinA) ( secA- cosA) ( tanA + cotA) = 1 prove ​

Answer 1

Step-by-step

• (cosecA−sinA)(secA−cosA)(tanA+cotA)
• (cosecA−sinA)(secA−cosA)(tanA+cotA)=(sinA1−sinA)(cosA1−cosA)(cosAsinA+sinAcosA)
• (cosecA−sinA)(secA−cosA)(tanA+cotA)=(sinA1−sinA)(cosA1−cosA)(cosAsinA+sinAcosA)=sinA(1−sin2A)×cosA(1−cos2A)×cosAsinAsin2A+cos2A
• (cosecA−sinA)(secA−cosA)(tanA+cotA)=(sinA1−sinA)(cosA1−cosA)(cosAsinA+sinAcosA)=sinA(1−sin2A)×cosA(1−cos2A)×cosAsinAsin2A+cos2A=sinA(cos2A)×cosA(sin2A)×cosAsinA1=1
• (cosecA−sinA)(secA−cosA)(tanA+cotA)=(sinA1−sinA)(cosA1−cosA)(cosAsinA+sinAcosA)=sinA(1−sin2A)×cosA(1−cos2A)×cosAsinAsin2A+cos2A=sinA(cos2A)×cosA(sin2A)×cosAsinA1=1(cosecA−sinA)(secA−cosA)(tanA+cotA)=1

Answer 2

Answer:

$:\implies \sf ( cosescA - sinA) \times ( secA- cosA) \times ( tanA + cotA) = 1$

Formulae need to know :

• ${\boxed{\sf{CosecA = \dfrac{1}{SinA}}}}$

• ${\boxed{\sf{SecA = \dfrac{1}{CosA}}}}$

• ${\boxed{\sf{TanA = \dfrac{SinA}{CosA}}}}$

• ${\boxed{\sf{Cot A = \dfrac{CosA}{SinA}}}}$

$:\implies \sf ( \dfrac{1}{SinA} - SinA ) \times ( \dfrac{1}{CosA} - CosA ) \times \dfrac{SinA}{CosA} + \dfrac{CosA}{SinA} = 1 </p><p></p><ul><li>[tex]{\boxed{\sf{1 - Sin^{2}A = Cos^{2}A}}}$

• ${\boxed{\sf{1 - Cos^{2}A = Sin^{2}A}}}$

• ${\boxed{\sf{Sin^{2}A + Cos^{2}A = 1 }}}$

$:\implies \sf ( \dfrac{1 - Sin^{2}A}{SinA} ) \times ( \dfrac{1 - Cos^{2}A}{CosA} ) \times \dfrac{Sin^{2}A + Cos^{2}A}{CosA \: SinA} = 1$

$:\implies \sf \dfrac{Cos^{2}A}{SinA} \times \dfrac{Sin^{2}A}{CosA} \times \dfrac{1}{CosA \: SinA} = 1$

$:\implies \sf ( \dfrac{Cos^{2}A \: Sin^{2}A }{Sin^{2}A \: Cos^{2}A} ) = 1$

LHS = RHS

${\boxed{\bf{Verified !}}}$