Answer
Verified
413.1k+ views
Hint: Use property of an isosceles triangle which is given as opposite angles of the two sides are equal if sides are equal. Now, use sine rule to find the value of the given expression. Sine rule is given as
$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$
Complete step-by-step answer:
Here, we have an isosceles triangle ABC, such that AB = AC = 10cm and BC = 18cm. And need to determine the value of ${{\tan }^{2}}C-{{\sec }^{2}}B+2=?$
We know the property of an isosceles triangle that the opposite angles of the equal opposite side are equal to each other.
As AB = AC =10cm, hence \[\angle B=\angle C\] by the above property.
Let $\angle B=\angle C=\theta $ and hence $\angle A$ is given as ${{180}^{o}}-2\theta $ by the property of triangle that sum of interior angles of a triangle is ${{180}^{\circ }}.$ Hence we get
$\begin{align}
& \angle B=\angle C=\theta \\
& \Rightarrow \angle A={{180}^{o}}-2\theta \\
\end{align}$
AB = AC = 10cm, BC = 18cm
The diagram for the triangle is as shown below:
So, we can apply ‘sine rule’ to calculate the given expression. Sine rule of any triangle ABC is given as
$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}...........\left( i \right)$
Here a, b, c are opposite sides to the angles $\angle A,\angle B,\angle C.$
Hence, from the equation (i), we get
$\dfrac{\sin \left( {{180}^{o}}-2\theta \right)}{18}=\dfrac{\sin \theta }{10}=\dfrac{\sin \theta }{10}$
$\Rightarrow \dfrac{\sin \left( {{180}^{o}}-2\theta \right)}{18}=\dfrac{\sin \theta }{10}................\left( ii \right)$
Now, we can replace $\sin \left( {{180}^{\circ }}-2\theta \right)$ by $\sin 2\theta $ by the identity, $\sin \left( 180-\theta \right)=\sin \theta $.
Hence, we get equation (ii) as,
$\dfrac{\sin 2\theta }{18}=\dfrac{\sin \theta }{10}$
Now, use trigonometric identity $\sin 2\theta =2\sin \theta \cos \theta $. Hence, we get
$\dfrac{2\sin \theta \cos \theta }{18}=\dfrac{\sin \theta }{10}$
Cancelling the like terms, we get
$\dfrac{2\cos \theta }{18}=\dfrac{1}{10}$
$\Rightarrow \cos \theta =\dfrac{9}{10}.................\left( iii \right)$
Now, replace $\theta $ by B, we get
$\cos B=\dfrac{9}{10}$
Now we know $\sec \theta =\dfrac{1}{\cos \theta }$ , so we get
$\sec B=\dfrac{10}{9}..................\left( iv \right)$
Now we know by Pythagoras theorem,
${{\left( Hypotenuse \right)}^{2}}={{(base)}^{2}}+{{(perpendicular)}^{2}}.........(v)$
Now we know, $\cos B=\dfrac{9}{10}=\dfrac{base}{hypotenuse}$, substituting these values in equation (v), we get
$\begin{align}
& {{\left( 10 \right)}^{2}}={{(9)}^{2}}+{{(perpendicular)}^{2}} \\
& \Rightarrow perpendicular=\sqrt{100-81}=\sqrt{19} \\
\end{align}$
So, we can write,
$\tan B=\dfrac{\text{Perpendicular}}{\text{Base}}=\dfrac{\sqrt{19}}{9}...................\left( vi \right)$
Now, we can solve the expression ${{\tan }^{2}}C-{{\sec }^{2}}B+2.$ Let it’s value be ‘M’. Hence we get,
$M={{\tan }^{2}}C-{{\sec }^{2}}B+2$
Now, we know that $\angle B=\angle C$ (isosceles triangle), so we can write
$M={{\tan }^{2}}B-{{\sec }^{2}}B+2$
Now, put values of $\tan B$ and $\sec B$ from the equation (iv) and (vi) hence, we get
$M={{\left( \dfrac{\sqrt{19}}{9} \right)}^{2}}-{{\left( \dfrac{10}{9} \right)}^{2}}+2$
$\Rightarrow M=\dfrac{19}{81}-\dfrac{100}{81}+2$
$\begin{align}
& \Rightarrow M=\dfrac{19-100}{81}+2 \\
& \Rightarrow M=\dfrac{-81}{81}+2=-1+2=1 \\
\end{align}$
So, the value of ${{\tan }^{2}}C-{{\sec }^{2}}B+2$ is 1.
Note: One can calculate value of $\tan C$and $\sec B$ , using the cosine formula as well which is given as, $\cos \theta =\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$
Here $\theta $ is the angle between side ’b’ and ‘c’.
Another approach is when we find the value of $\sec \theta $ , we can calculate the value of $\tan \theta $ using the formula, ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $.
$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$
Complete step-by-step answer:
Here, we have an isosceles triangle ABC, such that AB = AC = 10cm and BC = 18cm. And need to determine the value of ${{\tan }^{2}}C-{{\sec }^{2}}B+2=?$
We know the property of an isosceles triangle that the opposite angles of the equal opposite side are equal to each other.
As AB = AC =10cm, hence \[\angle B=\angle C\] by the above property.
Let $\angle B=\angle C=\theta $ and hence $\angle A$ is given as ${{180}^{o}}-2\theta $ by the property of triangle that sum of interior angles of a triangle is ${{180}^{\circ }}.$ Hence we get
$\begin{align}
& \angle B=\angle C=\theta \\
& \Rightarrow \angle A={{180}^{o}}-2\theta \\
\end{align}$
AB = AC = 10cm, BC = 18cm
The diagram for the triangle is as shown below:
So, we can apply ‘sine rule’ to calculate the given expression. Sine rule of any triangle ABC is given as
$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}...........\left( i \right)$
Here a, b, c are opposite sides to the angles $\angle A,\angle B,\angle C.$
Hence, from the equation (i), we get
$\dfrac{\sin \left( {{180}^{o}}-2\theta \right)}{18}=\dfrac{\sin \theta }{10}=\dfrac{\sin \theta }{10}$
$\Rightarrow \dfrac{\sin \left( {{180}^{o}}-2\theta \right)}{18}=\dfrac{\sin \theta }{10}................\left( ii \right)$
Now, we can replace $\sin \left( {{180}^{\circ }}-2\theta \right)$ by $\sin 2\theta $ by the identity, $\sin \left( 180-\theta \right)=\sin \theta $.
Hence, we get equation (ii) as,
$\dfrac{\sin 2\theta }{18}=\dfrac{\sin \theta }{10}$
Now, use trigonometric identity $\sin 2\theta =2\sin \theta \cos \theta $. Hence, we get
$\dfrac{2\sin \theta \cos \theta }{18}=\dfrac{\sin \theta }{10}$
Cancelling the like terms, we get
$\dfrac{2\cos \theta }{18}=\dfrac{1}{10}$
$\Rightarrow \cos \theta =\dfrac{9}{10}.................\left( iii \right)$
Now, replace $\theta $ by B, we get
$\cos B=\dfrac{9}{10}$
Now we know $\sec \theta =\dfrac{1}{\cos \theta }$ , so we get
$\sec B=\dfrac{10}{9}..................\left( iv \right)$
Now we know by Pythagoras theorem,
${{\left( Hypotenuse \right)}^{2}}={{(base)}^{2}}+{{(perpendicular)}^{2}}.........(v)$
Now we know, $\cos B=\dfrac{9}{10}=\dfrac{base}{hypotenuse}$, substituting these values in equation (v), we get
$\begin{align}
& {{\left( 10 \right)}^{2}}={{(9)}^{2}}+{{(perpendicular)}^{2}} \\
& \Rightarrow perpendicular=\sqrt{100-81}=\sqrt{19} \\
\end{align}$
So, we can write,
$\tan B=\dfrac{\text{Perpendicular}}{\text{Base}}=\dfrac{\sqrt{19}}{9}...................\left( vi \right)$
Now, we can solve the expression ${{\tan }^{2}}C-{{\sec }^{2}}B+2.$ Let it’s value be ‘M’. Hence we get,
$M={{\tan }^{2}}C-{{\sec }^{2}}B+2$
Now, we know that $\angle B=\angle C$ (isosceles triangle), so we can write
$M={{\tan }^{2}}B-{{\sec }^{2}}B+2$
Now, put values of $\tan B$ and $\sec B$ from the equation (iv) and (vi) hence, we get
$M={{\left( \dfrac{\sqrt{19}}{9} \right)}^{2}}-{{\left( \dfrac{10}{9} \right)}^{2}}+2$
$\Rightarrow M=\dfrac{19}{81}-\dfrac{100}{81}+2$
$\begin{align}
& \Rightarrow M=\dfrac{19-100}{81}+2 \\
& \Rightarrow M=\dfrac{-81}{81}+2=-1+2=1 \\
\end{align}$
So, the value of ${{\tan }^{2}}C-{{\sec }^{2}}B+2$ is 1.
Note: One can calculate value of $\tan C$and $\sec B$ , using the cosine formula as well which is given as, $\cos \theta =\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$
Here $\theta $ is the angle between side ’b’ and ‘c’.
Another approach is when we find the value of $\sec \theta $ , we can calculate the value of $\tan \theta $ using the formula, ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE