(1 + tan’0) (1 + coto)
1
(sin?0-sinºo)
Answers
Answer:
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Step-by-step explanation:
Answer:
Hence Proved
Step-by-step explanation:
Sin0(1+tan0)+cos0(1+cot0)=sec0+cosec0 prove this
\begin{gathered}Sin\theta(1+Tan\theta) + Cos\theta(1+Cot\theta)\\\\Tan\theta = \frac{Sin\theta}{Cos\theta} \: \& \:Cot\theta = \frac{Cos\theta}{Sin\theta}\\\\\implies Sin\theta(1+\frac{Sin\theta}{Cos\theta}) + Cos\theta(1+\frac{Cos\theta}{Sin\theta})\\\\\implies Sin\theta(\frac{Cos\theta + Sin\theta}{Cos\theta}) + Cos\theta(\frac{Sin\theta + Cos\theta}{Sin\theta})\\\\\end{gathered}
Sinθ(1+Tanθ)+Cosθ(1+Cotθ)
Tanθ=
Cosθ
Sinθ
&Cotθ=
Sinθ
Cosθ
⟹Sinθ(1+
Cosθ
Sinθ
)+Cosθ(1+
Sinθ
Cosθ
)
⟹Sinθ(
Cosθ
Cosθ+Sinθ
)+Cosθ(
Sinθ
Sinθ+Cosθ
)
\begin{gathered}\implies (Cos\theta + Sin\theta)(\frac{Sin\theta}{Cos\theta} + \frac{Cos\theta}{Sin\theta})\\\\\implies (Cos\theta + Sin\theta)(\frac{Sin^2\theta + Cos^2\theta}{Cos\theta Sin\theta})\\\\Sin^2\theta + Cos^2\theta = 1\\\\\implies (Cos\theta + Sin\theta)(\frac{1}{Cos\theta Sin\theta})\\\end{gathered}
⟹(Cosθ+Sinθ)(
Cosθ
Sinθ
+
Sinθ
Cosθ
)
⟹(Cosθ+Sinθ)(
CosθSinθ
Sin
2
θ+Cos
2
θ
)
Sin
2
θ+Cos
2
θ=1
⟹(Cosθ+Sinθ)(
CosθSinθ
1
)
\begin{gathered}\implies \frac{Cos\theta}{Cos\theta Sin\theta} + \frac{Sin\theta}{Cos\theta Sin\theta}\\\\\implies \frac{1}{Sin\theta} + \frac{1}{Cos\theta}\\\\\implies Cosec\theta + Sec\theta\\\\= RHS\end{gathered}
⟹
CosθSinθ
Cosθ
+
CosθSinθ
Sinθ
⟹
Sinθ
1
+
Cosθ
1
⟹Cosecθ+Secθ
=RHS
( PROVED )
.