# $\sin {{81}^{\circ }}+\tan {{81}^{\circ }}$, when expressed in terms of angles between ${{0}^{\circ }}$ and ${{45}^{\circ }}$, becomes

(a) $\sin {{9}^{\circ }}+\cos {{9}^{\circ }}$

(b) $cos{{9}^{\circ }}+\tan {{9}^{\circ }}$

(c) $\sin {{9}^{\circ }}+\tan {{9}^{\circ }}$

(d) $cos{{9}^{\circ }}+\cot {{9}^{\circ }}$

Last updated date: 27th Mar 2023

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Answer

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Hint: Think of changing or converting ‘sin’ to ‘cos’ and ‘tan’ to ‘cot’ respectively by using the identities $\sin \left( 90-\theta \right)=\cos \theta $ and $\tan \left( 90-\theta \right)=\cot \theta $. First try to prove it by using identities

\[\sin \left( A-B \right)\text{ }=\sin A\cos B\cos A\sin B\] and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$

Complete step-by-step answer:

In the given question we have to express $\left( \sin {{81}^{\circ }}+\tan {{81}^{\circ }} \right)$ in term of angles between ${{0}^{\circ }}$ and ${{45}^{\circ }}$ so that the value of expression also does not changes.

We will find out by using identity $\sin \left( 90-\theta \right)=\cos \theta $ and $\tan \left( 90-\theta \right)=\cot \theta $ to convert the angles between ${{0}^{\circ }}$and ${{45}^{\circ }}$ .

First we will prove the identity $\sin \left( 90-\theta \right)=\cos \theta $ by using formula \[\sin \left( A-B \right)\text{ }=\sin A\cos B\cos A\sin B\].

Now substituting \[A={{90}^{\circ }}\] and $B=\theta ,$ we get,

$\sin \left( 90-\theta \right)=\sin {{90}^{\circ }}\cos \theta -\cos 90\sin \theta $

Here we will put $\sin {{90}^{\circ }}=1$ and $\cos {{90}^{\circ }}=0$ we get,

$\sin \left( 90-\theta \right)=\cos \theta $

Now we will prove that identity $\tan \left( 90-\theta \right)=\cot \theta $

As we know that,

$\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha }$

So now we will replace $\alpha $ by $\left( 90-\theta \right)$ we get,

$\tan \left( 90-\theta \right)=\dfrac{\sin \left( 90-\theta \right)}{\cos \left( 90-\theta \right)}$

Now in this we will use identities $\sin \left( 90-\theta \right)=\cos \theta $and$\cos \left( 90-\theta \right)=\sin \theta $and substituting it we get,

$\tan \left( 90-\theta \right)=\dfrac{\cos \theta }{\sin \theta }=\cot \theta $

So now applying the identities we get,

$\begin{align}

& \sin \left( {{81}^{\circ }} \right)+\tan \left( {{81}^{\circ }} \right) \\

& =\sin \left( 90-9 \right)+\tan \left( 90-9 \right) \\

& =\cos 9+\cot 9. \\

\end{align}$

Hence, the expression $\sin \left( {{81}^{\circ }} \right)+\tan \left( {{81}^{\circ }} \right)$ can be expressed as $\cos {{9}^{\circ }}+\cot {{9}^{\circ }}$ for the angle to be in between ${{0}^{\circ }}$ and ${{45}^{\circ }}$.

So, the correct answer is option (d).

Note: Students are always in dilemma on how to approach these kinds of problems. They can do these kind of problems by using an easy method, that is just converting ‘sin’ to ‘cos’, ‘tan’ to ‘cot’ and ‘sec’ to ‘cosec’ ratios by using the identities $\sin \left( 90-\theta \right)=\cos \theta ,\tan \left( 90-\theta \right)=\cot \theta ,\sec \left( 90-\theta \right)=\csc \theta $ or vice-versa.

\[\sin \left( A-B \right)\text{ }=\sin A\cos B\cos A\sin B\] and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$

Complete step-by-step answer:

In the given question we have to express $\left( \sin {{81}^{\circ }}+\tan {{81}^{\circ }} \right)$ in term of angles between ${{0}^{\circ }}$ and ${{45}^{\circ }}$ so that the value of expression also does not changes.

We will find out by using identity $\sin \left( 90-\theta \right)=\cos \theta $ and $\tan \left( 90-\theta \right)=\cot \theta $ to convert the angles between ${{0}^{\circ }}$and ${{45}^{\circ }}$ .

First we will prove the identity $\sin \left( 90-\theta \right)=\cos \theta $ by using formula \[\sin \left( A-B \right)\text{ }=\sin A\cos B\cos A\sin B\].

Now substituting \[A={{90}^{\circ }}\] and $B=\theta ,$ we get,

$\sin \left( 90-\theta \right)=\sin {{90}^{\circ }}\cos \theta -\cos 90\sin \theta $

Here we will put $\sin {{90}^{\circ }}=1$ and $\cos {{90}^{\circ }}=0$ we get,

$\sin \left( 90-\theta \right)=\cos \theta $

Now we will prove that identity $\tan \left( 90-\theta \right)=\cot \theta $

As we know that,

$\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha }$

So now we will replace $\alpha $ by $\left( 90-\theta \right)$ we get,

$\tan \left( 90-\theta \right)=\dfrac{\sin \left( 90-\theta \right)}{\cos \left( 90-\theta \right)}$

Now in this we will use identities $\sin \left( 90-\theta \right)=\cos \theta $and$\cos \left( 90-\theta \right)=\sin \theta $and substituting it we get,

$\tan \left( 90-\theta \right)=\dfrac{\cos \theta }{\sin \theta }=\cot \theta $

So now applying the identities we get,

$\begin{align}

& \sin \left( {{81}^{\circ }} \right)+\tan \left( {{81}^{\circ }} \right) \\

& =\sin \left( 90-9 \right)+\tan \left( 90-9 \right) \\

& =\cos 9+\cot 9. \\

\end{align}$

Hence, the expression $\sin \left( {{81}^{\circ }} \right)+\tan \left( {{81}^{\circ }} \right)$ can be expressed as $\cos {{9}^{\circ }}+\cot {{9}^{\circ }}$ for the angle to be in between ${{0}^{\circ }}$ and ${{45}^{\circ }}$.

So, the correct answer is option (d).

Note: Students are always in dilemma on how to approach these kinds of problems. They can do these kind of problems by using an easy method, that is just converting ‘sin’ to ‘cos’, ‘tan’ to ‘cot’ and ‘sec’ to ‘cosec’ ratios by using the identities $\sin \left( 90-\theta \right)=\cos \theta ,\tan \left( 90-\theta \right)=\cot \theta ,\sec \left( 90-\theta \right)=\csc \theta $ or vice-versa.

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