Question

tan theta/tan(90°- theta) + sin(90°-theta)/cos theta=sec theta
prove
please help me ​

Answer 1

Answer:

To prove : \frac{\cos(90-\theta)\sec (90-\theta)\tan \theta }{\csc (90-\theta)\sin (90-\theta)\cot (90-\theta)}+\frac{\tan (90-\theta)}{\cot \theta}=2

csc(90−θ)sin(90−θ)cot(90−θ)

cos(90−θ)sec(90−θ)tanθ

+

cotθ

tan(90−θ)

=2

Proof :

Taking LHS,

=\frac{\cos(90-\theta)\sec (90-\theta)\tan \theta }{\csc (90-\theta)\sin (90-\theta)\cot (90-\theta)}+\frac{\tan (90-\theta)}{\cot \theta}=

csc(90−θ)sin(90−θ)cot(90−θ)

cos(90−θ)sec(90−θ)tanθ

+

cotθ

tan(90−θ)

Applying property of trigonometry,

\begin{gathered}\sin (90-\theta)=\cos \theta\\\cos(90-\theta)=\sin\theta\\\sec (90-\theta)=\csc\theta\\\csc (90-\theta)=\sec\theta\\\cot (90-\theta)=\tan\theta\\\tan (90-\theta)=\cot\theta\end{gathered}

sin(90−θ)=cosθ

cos(90−θ)=sinθ

sec(90−θ)=cscθ

csc(90−θ)=secθ

cot(90−θ)=tanθ

tan(90−θ)=cotθ

=\frac{\sin\theta\csc\theta\tan \theta }{\sec\theta\cos\theta\tan\theta}+\frac{\cot\theta}{\cot \theta}=

secθcosθtanθ

sinθcscθtanθ

+

cotθ

cotθ

We know, \csc\tehta=\frac{1}{\sin\theta},\ \sec\tehta=\frac{1}{\cos\theta}csc\tehta=

sinθ

1

, sec\tehta=

cosθ

1

=\frac{\sin\theta\times \frac{1}{\sin\theta}\times \tan \theta }{\frac{1}{\cos\theta}\times \cos\theta\times \tan\theta}+\frac{\cot\theta}{\cot \theta}=

cosθ

1

×cosθ×tanθ

sinθ×

sinθ

1

×tanθ

+

cotθ

cotθ

=\frac{1}{1}+1=

1

1

+1

=1+1=1+1

=2=2

LHS=RHS

Hence proved.

Answer 2

Answer:

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Step-by-step explanation:

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