Question

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1, tan 45° + 2 tan² 60° =

2, sin² 30° = 1 =

3, tan² 60° + 4 cos ² 45° + 3 sec² 30° + 5 cos² 90° = 

4, If tan $\large\theta = \frac{3}{4}$, then $\Large\ \frac{1 - cos\theta}{1 + cos\theta}$ =

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Answer 1

Step-by-step explanation:

$\LARGE\color{black}{\colorbox{#FF7968}{1.}}$ tan45⁰ + 2tan²60⁰ = ?

Using the standard values of trigonometric functions,

$\sf tan45⁰ = 1 ,\ tan60⁰ = \sqrt3$

Thus,

$\sf \bf :\longmapsto tan45⁰ + 2tan^260⁰ = 1 + (\sqrt3)^2$

$:\longmapsto\boxed{:\longmapsto\sf\bf tan45⁰ + 2tan^260⁰=4}$

$\LARGE\color{black}{\colorbox{#4FB3F6}{2.}}$ sin²30⁰+ 1 = ?

Using the standard values,

$\sf sin30⁰ = 1/2$

$\sf \bf :\longmapsto sin^230⁰ + 1 = \left(\dfrac{1}{2} \right)^2+ 1$

$:\longmapsto\boxed{\sf \bf sin^230⁰+1\dfrac{5}{4}}$

$\LARGE\color{black}{\colorbox{#FEDD8E}{3.}}$ tan²60⁰ + 4cos²45⁰ + 3sec²30⁰ + 5cos²90⁰ = ?

Using the standard values,

$\sf tan60⁰ = \sqrt3 , cos45⁰ = \dfrac{1}{2} , sec30⁰ = \dfrac{2}{\sqrt3}, cos90⁰ = 0$

$\sf \bf :\longmapsto tan^260⁰ + 4cos^245⁰ + 3sec^230⁰ + 5cos^290⁰ = (\sqrt3)^2+ 4\left(\dfrac{1}{2}\right)^2 + 3\left(\dfrac{2}{\sqrt3}\right)^2 +5(0)$

$\sf \bf :\longmapsto tan^260⁰ + 4cos^245⁰ + 3sec^230⁰ + 5cos^290⁰ = 3 + 2 + 4 + 0$

$:\longmapsto \boxed{\sf\bf tan^260⁰ + 4cos^245⁰ + 3sec^230⁰ + 5cos^290⁰ = 9}$

$\LARGE\color{black}{\colorbox{#FBBE2E}{4.}}$ $\theta = \frac{3}{4}$, then $\Large\ \frac{1 - cos\theta}{1 + cos\theta} = ?$

Consider the diagram for this part

As the ratio of AB and AC is 4/3 considering $\theta$$\angle$C,

Let AB and BC be 3k and 4k

Thus, AC² = AB² + BC² [Pythagoras theorem]

$\sf \bf :\longmapsto AC^2 = (3k)^2 + (4k)^2$

$\sf \bf :\longmapsto AC^2= 9k^2 + 16k^2$

$\sf \bf :\longmapsto AC^2 = 25k^2$

$\sf \bf :\longmapsto AC = 5k$

Now , clearly,

$\sf \bf :\longmapsto cos \theta = \displaystyle \frac{BC}{AC} = \frac{4k}{5k} = \frac{4}{5}$

Putting this value in the expression,

$\sf \bf :\longmapsto \dfrac{1 - cos\theta}{1 + cos\theta} = \dfrac{1-\dfrac{4}{5}}{1+\dfrac{4}{5}}$

$\sf \bf :\longmapsto \dfrac{1 - cos\theta}{1 + cos\theta} = \dfrac{1}{5} \times \dfrac{5}{9}$

$:\longmapsto \boxed{\sf \bf \dfrac{1 - cos\theta}{1 + cos\theta} = \dfrac{1}{9}}$

Answer 2

Answer:

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