prove that cos(90- theta)- sin (90-theta) ( cosec theta- sin theta) ( tan theta+ cot theta) = 1 About the author Samantha
Step-by-step explanation: Answer and explanation: 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 Reply
Step-by-step explanation:
Answer and explanation:
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