Question

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Answer 1

Answer:—

$\sf{tan\theta=\sqrt{2}-1}$

Step-by-step explanation:

$\sf{Given\colon\:sin\theta+cos\theta=\sqrt{2}cos\theta}$

$\sf{sin\theta=\sqrt{2}cos\theta-cos\theta}$

$\sf{sin\theta=cos\theta\left(\sqrt{2}-1\right)}$

$\sf{\dfrac{sin\theta}{cos\theta}=\sqrt{2}-1}$

$\sf{we\:know\:that\:\dfrac{sin\theta}{cos\theta}=tan\theta}$

$\boxed{\sf{tan\theta=\sqrt{2}-1}}$

Answer 2

$\huge\mathcal{\mid{\mid{\underline{\purple{Good\: Afternoon\:}}}{\mid{\mid}}}}$

Qᴜᴇsᴛɪᴏɴ ;-

❥︎ If $\bf{\sin\theta\:+\:\cos\theta\:=\:\sqrt{2}\:\cos\theta\:}$, then find the value of $\bf{\tan\theta}$ .

$\huge{\orange{\boxed{\fcolorbox{cyan}{indigo}{\color{lime}ANSWER}}}}$

$\pink\longmapsto\:\:\bf\blue{\sin\theta\:+\:\cos\theta\:=\:\sqrt{2}\:\cos\theta\:}$

$\longmapsto\:\:\sf\pink{\sin\theta\:=\:\sqrt{2}\:\cos\theta\:-\:\cos\theta\:}$

$\longmapsto\:\:\sf\orange{\sin\theta\:=\:\Big(\sqrt{2}\:-\:1\Big)\:\cos\theta\:}$

$\longmapsto\:\:\sf\purple{\dfrac{\sin\theta}{\cos\theta}\:=\:\Big(\sqrt{2}\:-\:1\Big)\:}$

$\Large\bf\pink{We\: know\: that},$

• $\bf\red{\dfrac{\sin\theta}{\cos\theta}\:=\:\tan\theta}$

$\pink\longmapsto\:\:\bf\green{\tan\theta\:=\:\Big(\sqrt{2}\:-\:1\Big)\:}$