JEE Main 2021 — Mathematics Algebra

Let $f(x)=\mathrm{cos}\lambda x$f(x)=cosλx

$$\because f(\frac{1}{2})=-1$$∵f(21​)=−1

So, $-1=\mathrm{cos}\frac{\lambda }{2}\Rightarrow \frac{\lambda }{2}=\pi$−1=cos2λ​⇒2λ​=π

Thus $f(x)=\mathrm{cos}2\pi x$f(x)=cos2πx

Now $k$k is a natural number

Thus $f(k)=1$f(k)=1

$$\sum _{k=1}^{20}\frac{1}{\mathrm{sin}(k)\mathrm{sin}(k+1)}=\frac{1}{\mathrm{sin}1}\sum _{k=1}^{20}[\frac{\mathrm{sin}((k+1)-k)}{\mathrm{sin}k\cdot \mathrm{sin}(k+1)}]$$k=1∑20​sin(k)sin(k+1)1​=sin11​k=1∑20​[sink⋅sin(k+1)sin((k+1)−k)​]

$$\frac{1}{\mathrm{sin}1}\sum _{k=1}^{20}[\frac{\mathrm{sin}(k+1)\mathrm{cos}k-\mathrm{sin}k\mathrm{cos}(k+1)}{\mathrm{sin}k\cdot \mathrm{sin}(k+1)}]$$sin11​k=1∑20​[sink⋅sin(k+1)sin(k+1)cosk−sinkcos(k+1)​]

$$=\frac{1}{\mathrm{sin}1}\sum _{k=1}^{20}(\mathrm{cot}k-\mathrm{cot}(k+1))$$=sin11​k=1∑20​(cotk−cot(k+1))

$$=\frac{\mathrm{cot}1-\mathrm{cot}21}{\mathrm{sin}1}=\frac{\frac{\mathrm{cos}1}{\mathrm{sin}1}-\frac{\mathrm{cos}21}{\mathrm{sin}21}}{\mathrm{sin}1}$$=sin1cot1−cot21​=sin1sin1cos1​−sin21cos21​​

$$=\frac{\mathrm{sin}21\mathrm{cos}1-\mathrm{sin}1\mathrm{cos}21}{{\mathrm{sin}}^{2}1\mathrm{sin}21}=\mathrm{sin}(21-1){\mathrm{cosec}}^{2}1\mathrm{cosec}21$$=sin21sin21sin21cos1−sin1cos21​=sin(21−1)cosec21cosec21

$$={cosec}^{2}(1)cosec(21)\mathrm{sin}(20)$$=cosec2(1)cosec(21)sin(20)