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Solve and prove the following identities: (v) $(1+cotA-cosecA)(1+tanA+secA)=2$ (vi) $sinAcotA+sinAcosecA=1+cosA$
Solve and prove the following identities: (v) $(1+cotA-cosecA)(1+tanA+secA)=2$ (vi) $sinAcotA+sinAcosecA=1+cosA$
Class 10 | Frank | ICSE | Maths | Trigonometric Identities
Solve and prove the following identities: (v) $(1+cotA-cosecA)(1+tanA+secA)=2$ (vi) $sinAcotA+sinAcosecA=1+cosA$

(v)

From the question firstly we consider Left Hand Side (LHS),

$(1+cotA–cosecA)(1+tanA+secA)$

We know that,

$cotA=sinA/cosA$, $cosecA=1/cosA$, $tanA=cosA/sinA$, $secA=1/sinA$

$(1+(sinA/cosA)+(1/cosA))(1+(cosA/sinA)–(1/sinA))$

Taking LCM we get,

$((cosA+sinA+1)/cosA)((sinA+cosA–1)/sinA)$

$\left( {{\left( \sin A+\cos A \right)}^{2}}-{{1}^{2}} \right)/\left( \sin A\cos A \right)$

$=(1+2sinAcosA–1)/(sinAcosA)$

$=(2sinAcosA)/(sinAcosA)$

By simplification we get,

$=2$

Then, Right Hand Side (RHS) $=2$

Therefore, LHS = RHS

(vi)

From the question firstly we consider Left Hand Side (LHS),

= $sinAcotA+sinAcosecA$

We know that, $cotA=cosA/sinA$, $cosecA=1/sinA$

= $sinA(cosA/sinA)+sinA(1/sinA)$

= $cosA+1$

Then, Right Hand Side (RHS) $=1+cosA$

Therefore, LHS = RHS

i

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