Question

if sin theta= 4/5 then find 1- tan theta /1+tan theta ​

Answer 1

Answer:

$\sf{The \ value \ of \ \dfrac{1-tan\theta}{1+tan\theta} \ is \ \dfrac{-1}{7}.}$

Given:

• $\sf{sin\theta=\dfrac{4}{5}}$

To find:

• $\sf{The \ value \ of \ \dfrac{1-tan\theta}{1+tan\theta}}$

Solution:

$\sf{sin^{2}\theta+cos^{2}\theta=1}$

$\sf{... Trigonometric \ identity}$

$\sf{\therefore{cos^{2}\theta=1-sin^{2}\theta}}$

$\sf{\therefore{cos^{2}\theta=1-(\dfrac{4}{5})^{2}}}$

$\sf{\therefore{cos^{2}\theta=1-\dfrac{16}{25}}}$

$\sf{\therefore{cos^{2}\theta=\dfrac{25-16}{25}}}$

$\sf{\therefore{cos^{2}\theta=\dfrac{9}{25}}}$

$\sf{\therefore{cos\theta=\dfrac{3}{5}}}$

$\sf{tan\theta=\dfrac{sin\theta}{cos\theta}}$

$\sf{\therefore{tan\theta=\dfrac{\frac{4}{5}}{\frac{3}{5}}}}$

$\sf{\therefore{tan\theta=\dfrac{4}{3}}}$

$\sf{\leadsto{\dfrac{1-tan\theta}{1+\tan\theta}}}$

$\sf{\leadsto{\dfrac{1-\frac{4}{3}}{1+\frac{4}{3}}}}$

$\sf{\leadsto{\dfrac{\frac{3-4}{3}}{\frac{3+4}{3}}}}$

$\sf{\leadsto{\dfrac{-1}{3}\times\dfrac{3}{7}}}$

$\sf{\leadsto{\dfrac{-1}{7}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ \dfrac{1-tan\theta}{1+tan\theta} \ is \ \dfrac{-1}{7}.}}}$

Answer 2

$\bf\huge\blue{\underline{\underline{ Question : }}}$

If sin θ = 4/5, then find 1 - tan θ/1 + tan θ

$\bf\huge\blue{\underline{\underline{ Solution : }}}$

★ Given that,

• sin θ = 4/5

★ To find,

• 1 - tan θ/1 + tan θ

★ Let,

We know that,

$\tt \rightarrow \sin \theta = \cfrac{ Opposite}{ Hypotenuse } = \cfrac{4}{5}$

So,

• AC = 5 cm.
• BC = 4 cm.
• AB = ?

By using Pythagoras Theorem, we get the value of AC.

Pythagoras Theorem :

➡ (Hypotenuse)² = (Side)² + (Side)²

$\sf \leadsto (AC)^{2} = (AB)^{2} + (BC)^{2}$

$\sf \leadsto (5)^{2} = (AB)^{2} + (4)^{2}$

$\sf \leadsto 25 = (AB)^{2} + 16$

$\sf \leadsto (AB)^{2} = 25-16$

$\sf \leadsto (AB)^{2} = 9$

$\sf \leadsto (AB)= \sqrt{9}$

$\sf \leadsto (AB)= 3$

★ Now,

We can find out the value of 1 - tan θ/1 + tan θ

$\sf \implies \tan\:\theta = \cfrac{ Opposite}{ Adjacent }$

$\sf \implies \tan\:\theta = \cfrac{ BC}{AB}$

$\sf \implies \tan\:\theta = \cfrac{ 4}{3}$

• Now, substitute value of tan θ.

$\sf \implies \cfrac{ 1 - \frac{4}{3}}{1+\frac{4}{3}}$

$\sf \implies \cfrac{\frac{3 - 4}{3}}{\frac{3+4}{3}}$

$\sf \implies \cfrac{-1}{\cancel{3}} \times \cfrac{\cancel{3}}{7}$

$\sf \implies \cfrac{-1}{7}$

$\underline{\boxed{\rm{\purple{\therefore Hence,\:the\:value\:of\:\cfrac{1 - \tan\:\theta}{1+\tan\:\theta}=\cfrac{-1}{7}.}}}}\:\orange{\bigstar}$

★ More Information,

$\boxed{\begin{minipage}{7 cm} Trigonometric Identities : \\ \\ \sin^{2}\theta + cos^{2}\theta = 1 \\ \\ 1 + tan^{2}\theta = sec^{2}\theta \\ \\1 + cot^{2}\theta=\text{cosec}^2\, \theta \end{minipage}}$

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