Question

5. fH
: (i) sin A (1 + tan A) + cos A (1 + cot A) = sec A+ cosec A
sin 0-cose +1
sin 0+ cos 0-1 sec 0 - tano​

Answer 1

Question: Prove that sinA (1 + tanA) + cosA (1 + cotA) = secA + cosecA.

Solution: sinA (1 + tanA) + cosA (1 + cotA) = secA + cosecA

• 1/tanA = cotA

L.H.S → sinA (1 + tanA) + cosA (1 + 1/tanA)

→ sinA (1 + tanA) + cosA (tanA + 1)/tanA

→ sinA (1 + tanA) + cosA/tanA (1 + tanA)

→ (1 + tanA)(sinA + cosA/tanA)

• tanA = sinA/cosA

→ (1 + tanA)(sinA + cosA × cosA/sinA)

→ (1 + tanA)(sinA + cos²A/sinA)

→ (1 + tanA)(sin²A + cos²A)/sinA

→ (1 + tanA)/sinA

→ 1/sinA + tanA/sinA

→ 1/sinA + sinA/cosA × 1/sinA

→ 1/sinA + 1/cosA

• 1/cosA = secA
• 1/sinA = cosecA

→ cosecA + secA

Q.E.D

Answer 2

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⠀

$\implies\sin( \alpha ) (1 + \tan( \alpha )) + \cos( \alpha ) (1 + \cot( \alpha ) )$

$\implies( \alpha ) (1 + \dfrac{ \sin( \alpha ) }{ \cos( \alpha ) } ) + \cos( \alpha ) (1 + \dfrac{ \cos( \alpha ) }{ \sin( \alpha ) } )$

$\implies\sin( \alpha )( \dfrac{ \cos( \alpha ) + \sin( \alpha ) }{ \cos( \alpha ) }) + \cos( \alpha ) ( \dfrac{ \sin( \alpha ) + \cos( \alpha ) }{ \sin( \alpha ) })$

$\implies\dfrac{ \sin( \alpha ) ( \cos( \alpha ) + \sin( \alpha ) )}{ \cos( \alpha ) } + \dfrac{ \cos( \alpha ) (\sin( \alpha ) + \cos( \alpha )) }{ \sin( \alpha ) }$

$\implies( \sin( \alpha ) + \cos( \alpha )) ( \dfrac{ \sin( \alpha ) }{ \cos( \alpha ) } + \dfrac{ \cos( \alpha ) }{ \sin( \alpha ) })$

$\implies( \sin( \alpha ) + \cos( \alpha )) \times \dfrac{ \sin^{2} ( \alpha ) + { \cos }^{2} (\alpha )}{( \cos( \alpha ) \sin( \alpha ) )}$

$\implies( \sin( \alpha ) + \cos( \alpha )) \times \dfrac{ 1}{( \cos( \alpha ) \sin( \alpha ) )}$

$\implies\dfrac{ ( \sin( \alpha ) + \cos( \alpha ))}{( \cos( \alpha ) \sin( \alpha ) )}$

$\implies\dfrac{ \cancel{\sin( \alpha ) }}{( \cos( \alpha ) \cancel{\sin( \alpha )} )} + \dfrac{ \cancel{\cos( \alpha )} }{( \cancel{\cos( \alpha )} \sin( \alpha ) )}$

$\implies \dfrac{1}{ \cos( \alpha ) } + \dfrac{1}{ \sin( \alpha ) }$

$\implies \sec( \alpha ) + \csc( \alpha )$