Question

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Answer 1

sin θ ( 1 + tan θ ) + cos θ ( 1 + cot θ )

sin θ ( 1 + sin θ / cos θ ) + cos θ ( 1 + cos θ / sin θ )

sin θ ( cos θ + sin θ ) / cos θ + cos θ ( sin θ + cos θ ) / sin θ

( cos θ + sin θ ) ( tan θ + cot θ )

( cos θ + sin θ ) ( sin² θ + cos²θ ) / ( sin θ cos θ )

( cos θ + sin θ ) / ( sin θ cos θ ) since sin² θ + cos²θ = 1

[ cos θ / ( sin θ cos θ ) ] + [ sin θ / ( sin θ cos θ ) ]

= 1 / sin θ + 1 / cos θ

Answer 2

▶ Question :-

Prove that :

→ sin∅( 1 + tan ∅ ) + cos∅( 1 + cot ∅ ) = ( sec∅ + cosex∅ ) .

▶ Step-by-step explanation :-

Solving LHS :-)

$\sf \because \sin \theta(1 + \tan \theta) + \cos \theta(1 + \cot \theta) \\ \\ \sf = \sin \theta + \sin \theta \times \frac{ \sin \theta}{ \cos \theta} + \cos \theta + \cos \theta \times \frac{ \cos \theta}{ \sin \theta} . \\ \\ = \frac{ \sin^{2}\theta \cos \theta + { \sin}^{2} \theta + { \cos}^{2} \theta \sin \theta + { \cos}^{2} \theta }{ \sin \theta \cos \theta} \\ \\ = \frac{( { \sin}^{3} \theta + { \cos}^{3} \theta) + ( \cos \theta { \sin}^{2} \theta + { \cos}^{2} \theta \sin \theta)}{ \cos \theta \sin \theta } . \\ \\ = \frac{( \sin \theta + \cos \theta)( { \sin}^{2} \theta - \sin \theta \cos \theta + { \cos}^{2} \theta) + \sin \theta \cos \theta( \sin \theta + \cos \theta)}{ \cos \theta \sin \theta} . \\ \\ = \frac{( \sin \theta + \cos \theta)( { \sin}^{2} \theta + { \cos}^{2} \theta \cancel{ - \sin \theta \cos \theta} \cancel{ + \sin \theta \cos \theta})}{ \cos \theta \sin \theta} . \\ \\ = \frac{( \sin \theta + \cos \theta) \times 1.}{ \cos \theta \sin \theta} . \\ \\ = \frac{ \cancel{ \sin \theta}}{ \cos \theta \cancel{\sin \theta}} + \frac{ \cancel{\cos \theta}}{ \cancel{ \cos \theta} \sin \theta} . \\ \\ = \frac{1}{ \cos \theta} + \frac{1}{ \sin \theta} . \\ \\ \huge \pink{ \boxed{ \boxed{ \tt = \sec \theta + cosec \theta.}}}$

✔✔ Hence, it is proved ✅✅.

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