Question
Tan θ = 1.Find Sin θ= , Cos θ= ?
✓
Answer
$\tan \theta=1=\frac{1}{1}$..(i) [Given]
In right angled $\triangle ABC$,
$\angle C=\theta \text {. }$
$\tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}$
$\therefore \quad \tan \theta=\frac{A B}{B C}$
$\therefore \quad \frac{ AB }{ BC }=\frac{1}{1}$
Let the common multiple be k.
$\therefore AB = 1k and BC = 1k$
Now, $AC^2 = AB^2 + BC^2 …[Pythagoras theorem]$
$= K^2 + K^2$
$= 2K^2$
$\therefore AC$ = $\sqrt { 2 k }$
$ \therefore \quad \sin \theta =\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{ lk }{\sqrt{2} k }=\frac{1}{\sqrt{2}}$
$\cos \theta =\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{ lk }{\sqrt{2} k }=\frac{1}{\sqrt{2}}$
In right angled $\triangle ABC$,
$\angle C=\theta \text {. }$
$\tan \theta=\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}$
$\therefore \quad \tan \theta=\frac{A B}{B C}$
$\therefore \quad \frac{ AB }{ BC }=\frac{1}{1}$
Let the common multiple be k.
$\therefore AB = 1k and BC = 1k$
Now, $AC^2 = AB^2 + BC^2 …[Pythagoras theorem]$
$= K^2 + K^2$
$= 2K^2$
$\therefore AC$ = $\sqrt { 2 k }$
$ \therefore \quad \sin \theta =\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{ lk }{\sqrt{2} k }=\frac{1}{\sqrt{2}}$
$\cos \theta =\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{ lk }{\sqrt{2} k }=\frac{1}{\sqrt{2}}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Factorise:
$x^2 + 2xy + y^2 - a^2 + 2ab - b^2$
→$x^2 + 2xy + y^2 - a^2 + 2ab - b^2$
The cost of ball pen is Rs. 5 less than half of the cost of fountain pen. Write this statement as a linear equation in two variables.
Factorise:
$x^4 y^4-x y$
→$x^4 y^4-x y$
The mean weight of 8 numbers is 15. If each number is multiplied by 2, what will be the new mean?
Write the following in decimal form and say what kind of decimal expansion has.$\frac{3}{11}$
→Two supplementary angles differ by 48°. Find the angles.
Factorize the following expressions:
$(a-3 b)^3+(3 b-c)^3+(c-a)^3$
→$(a-3 b)^3+(3 b-c)^3+(c-a)^3$
If a and b are distinct primes such that $\sqrt[3]{\text{a}^6\text{b}^{-4}}=\text{a}^\text{x}\text{b}^{2\text{y}},$find x and y.
→Factorise:
$9 x^2-3 x-20$
→$9 x^2-3 x-20$
Write the complement of an angle of measure x°.