Question

# $\sin {{81}^{\circ }}+\tan {{81}^{\circ }}$, 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) $cos{{9}^{\circ }}+\cot {{9}^{\circ }}$

Hint: Think of changing or converting â€˜sinâ€™ to â€˜cosâ€™ and â€˜tanâ€™ to â€˜cotâ€™ 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 $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$

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 â€˜sinâ€™ to â€˜cosâ€™, â€˜tanâ€™ to â€˜cotâ€™ and â€˜secâ€™ to â€˜cosecâ€™ 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.